Rd Sharma 2020 Solutions for Class 10 Maths Chapter 8 Circles are provided here with simple step-by-step explanations. These solutions for Circles are extremely popular among Class 10 students for Maths Circles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2020 Book of Class 10 Maths Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2020 Solutions. All Rd Sharma 2020 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

#### Page No 8.33:

#### Question 1:

If *PT* is a tangent at T to a circle whose centre is *O* and *OP *= 17 cm, *OT* = 8 cm, Find the length of the tangent segment *PT*.

#### Answer:

Let us put the given data in the form of a diagram.

We have to find TP. From the properties of tangents we know that a tangent will always be at right angles to the radius of the circle at the point of contact. Therefore is a right angle and triangle *OTP* is a right triangle.

We can find the length of *TP* using Pythagoras theorem. We have,

Therefore, the length of* TP* is 15 cm.

#### Page No 8.33:

#### Question 2:

Find the length of a tangent drawn to a circle with radius 5 cm, from a point 13 cm from the centre of the circle.

#### Answer:

Let us first put the given data in the form of a diagram.

We have to find TP. From the properties of tangents we know that a tangent will always be at right angles to the radius of the circle at the point of contact. Therefore is a right angle and triangle *OTP* is a right triangle.

We can find the length of *TP* using Pythagoras theorem. We have,

Therefore, the length of* TP* is 12 cm.

#### Page No 8.33:

#### Question 3:

*A* point *P* is 26 cm away from the centre *O* of a circle and the length *PT* of the tangent drawn from *P* to the circle is 10 cm. Find the radius of the circle.

#### Answer:

Let us put the given data in the form of a diagram.

We have to find *OT*. From the properties of tangents we know that a tangent will always be at right angles to the radius of the circle at the point of contact. Therefore is a right angle and triangle *OTP* is a right triangle.

We can find the length of *TP* using Pythagoras theorem. We have,

Therefore, the radius of the circle is 24 cm.

#### Page No 8.33:

#### Question 4:

If from any point on the common chord of two interesting circle, tangents be drawn to the circles, prove that they are equal.

#### Answer:

Let the two circles intersect at points X and Y. XY is the common chord.

Suppose A is a point on the common chord and AM and AN be the tangents drawn from A to the circle.

We need to show that AM = AN.

In order to prove the above relation, following property will be used.

“Let PT be a tangent to the circle from an external point P and a secant to the circle through P intersects the circle at points A and B, then PT ^{2 }= PA × PB”.

Now, AM is the tangent and AXY is a secant.

∴ AM^{2} = AX × AY ...(1)

AN is the tangent and AXY is a secant.

∴ AN^{2} = AX × AY ...(2)

From (1) and (2), we have

AM^{2}^{ }= AN^{2}

∴ AM = AN

#### Page No 8.33:

#### Question 5:

If the sides of a quadrilateral touch a circle. prove that the sum of a pair of opposite sides is equal to the sum of the other pair.

#### Answer:

Let us first put the given data in the form of a diagram.

We have been asked to prove that the sum of the pair of opposite sides of the quadrilateral is equal to the sum of the other pair.

Therefore, we shall first consider,

*AB + DC*

But by looking at the figure we have,

*AB + DC = AF + FB + DH + HC* …… (1)

From the property of tangents we know that the length of two tangents drawn to a circle from a common external point will be equal. Therefore we have the following,

*AF = AE*

*FB = BG*

*DH = ED*

*HC = CG*

Replacing for all the above in equation (1), we have

*AB + DC = AE + BG + ED + CG*

*AB + DC = (AE + ED) + (BG +CG)*

*AB + DC = AD + BC*

Thus we have proved that the sum of the pair of opposite sides of the quadrilateral is equal to the sum of the other pair.

#### Page No 8.33:

#### Question 6:

Out of the two concentric circles , the radius of the outer circle is 5 cm and the chord* AC *of length 8 cm is a tangent to the inner circle . Find the radius of the inner circle .

#### Answer:

Let the centre of the two concentric circles be O.

CD is the tangent to the inner circle.

OP joins the centre of the circle to the tangent at the point of contant.

OP $\perp $ CD and OP bisects CD.

Thus, PD = 4 cm

In ∆OPD,

${\mathrm{OP}}^{2}+{\mathrm{PD}}^{2}={\mathrm{OD}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{OP}}^{2}={\mathrm{OD}}^{2}-{\mathrm{PD}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{OP}}^{2}={\left(5\right)}^{2}-{\left(4\right)}^{2}=25-16=9\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OP}=3\mathrm{cm}$

Hence, the radius of the inner circle = 3 cm.

#### Page No 8.33:

#### Question 7:

A chord *PQ *of a circle is parallel to the tangent drawn at a point *R* of the circle . Prove that* R *bisects the arc *PRQ*.

#### Answer:

**Given:** Circle with centre O. PQ is the chord parallel to the tangent *l* at R

**To prove:** The point R bisects the arc PRQ.

**Construction:** Join OR intersecting PQ at S.

**Proof:**

OR ⊥ *l* (Radius is perpendicular to the tangent at the point of contact)

PQ || *l* (given)

∴∠OSP = ∠OSQ = 90° (pair of corresponding angles)

In ΔOPS and ΔOQS

OP = OQ (Radii of the same circle)

OS = OS (Common)

∠OSP = ∠OSQ (Proved)

So,ΔOPS ≅ ΔOQS (RHS congruence criterion)

⇒ ∠POS = ∠QOS (C.P.C.T)

⇒ arc (PR) = arc (QR) (Measure of the arc is same as the angle subtended by the arc at the centre)

Thus, the point R bisects the arc (PRQ).

#### Page No 8.33:

#### Question 8:

Prove that a diameter *AB *of a circle bisects all those chords which are parallel to the tangent at that point *A* .

#### Answer:

*l*is tangent that touches the circle at painted to the circle at point A. Let AB is diameter. Consider a chord CD parallel to

*l*. AB intersects CD at point Q.

*l*is tangent to the circle at point A.

*l*

*l*|| CD

*l*.

#### Page No 8.33:

#### Question 9:

If AB, AC, PQ are tangents in the given figure and AB = 5 cm, find the perimeter of Δ APQ.

#### Answer:

We have been asked to find the perimeter of the triangle *APQ*.

Therefore,

Perimeter of Δ*APQ* is equal to* AP + AQ + PQ*

By looking at the figure, we can rewrite the above as follows,

Let the Perimeter of Δ*APQ *be *P*. So *P*= *AP + AQ + PX + XQ*

From the property of tangents we know that when two tangents are drawn to a circle from the same external point, the length of the two tangents will be equal. Therefore we have,

*PX =PB*

*XQ =QC*

Replacing these in the above equation we have,

*P *=*AP + AQ + PB + QC*

From the figure we can see that,

*AP + PB = AB*

*AQ + QC = AC*

Therefore, we have, *P*= *AB + AC*

It is given that *AB* = 5 cm.

Again from the same property of tangents we know that that when two tangents are drawn to a circle from the same external point, the length of the two tangents will be equal. Therefore we have,

*AB = AC*

Therefore,

*AC* = 5 cm

Hence,

*P *= 5 + 5=10

Thus the perimeter of triangle *APQ* is 10 cm.

#### Page No 8.34:

#### Question 10:

Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre.

#### Answer:

Given: *XY* and *X*′*Y*′ at are two parallel tangents to the circle with centre *O* and *AB* is the tangent at the point *C*, which intersects *XY* at *A* and *X*′*Y*′ at B.

To Prove: ∠*AOB* = 90°

Construction: Let us joint point *O* to *C*.

Proof:

In Δ*OPA* and Δ*OCA*, we have

*OP = OC* (Radii of the same circle)

*AP = AC* (Tangents from point *A*)

*AO = AO* (Common side)

Δ*OPA* ≅ Δ*OCA* (SSS congruence criterion)

Therefore, ∠*POA* = ∠*COA* ……(*i*) (C.P.C.T)

Similarly, Δ*OQB* ≅ Δ*OCB* ……(*ii*)

Since *POQ* is a diameter of the circle, it is a straight line.

Therefore, ∠*POA* + ∠*COA* + ∠*COB* + ∠*QOB* = 180°

From equations (*i*) and (*ii*), it can be observed that

2∠*COA* + 2∠*COB* = 180°

∴ ∠*COA* + ∠*COB* = 90°

So, ∠AOB = 90°.

#### Page No 8.34:

#### Question 11:

In the given figure, *PQ *is tangent at a point* R* of the circle with centre *O*. If ∠*TRQ* = 30°, find *m* ∠*PRS*.

#### Answer:

We have been given that.

From the property of tangents we know that a tangent will always be perpendicular to the radius at the point of contact. Therefore,

Looking at the given figure we can rewrite the above equation as follows,

We know that. Therefore,

Now consider. The two sides of this triangle *OR* and *OT* are nothing but the radii of the same circle. Therefore,

*OR = OT*

And hence is an isosceles triangle. We know that the angles opposite to the equal sides of the isosceles triangle will be equal. Therefore,

We have found out that. Therefore,

=

Now consider. We know that sum of all angles of a triangle will always be equal to . Therefore,

Now let us consider the straight line *SOT*. We know that the angle of a straight line is . Therefore,

From the figure we can see that,

That is,

=

We have found out that. Therefore,

Let us take up now. The sides *SO* and *OR* of this triangle are nothing but the radii of the same circle and hence they are equal. Therefore,is an isosceles triangle. In an isosceles triangle, the angles opposite to the two equal sides of the triangle will be equal. Therefore we have,

Also the sum of all angles of a triangle will be equal to. Therefore,

In the previous step we have found out that. Therefore,

Let us now take up . We know from the property of tangents that the angle between the radius of the circle and the tangent at the point of contact will be equal to . Therefore,

=

By looking at the figure we can rewrite the above equation as follows,

In the previous section we have found that. Therefore,

Thus we have found out that .

#### Page No 8.34:

#### Question 12:

If *PA* and *PB* are tangents from an outside point *P* such that *PA* = 10 cm and ∠*APB* = 60°. Find the length of chord *AB*.

#### Answer:

Let us first put the given data in the form of a diagram.

From the property of tangents we know that the length of two tangents drawn to a circle from a common external point will always be equal. Therefore,

*PA=PB*

Consider the triangle *PAB*. Since we have *PA=PB*, it is an isosceles triangle. We know that in an isosceles triangle, the angles opposite to the equal sides will be equal. Therefore we have,

Also, sum of all angles of a triangle will be equal to . Therefore,

Since we know that,

Now if we see the values of all the angles of the triangle, all the angles measure. Therefore triangle *PAB* is an equilateral triangle.

We know that in an equilateral triangle all the sides will be equal.

It is given in the problem that side *PA* = 10 cm. Therefore, all the sides will measure 10 cm. Hence, *AB* = 10 cm.

Thus the length of the chord *AB* is 10 cm.

#### Page No 8.34:

#### Question 13:

In a right triangle *ABC* in which $\angle $*B*= 90^{0} , a circle is drawn with *AB *as diameter intersecting the hypotenuse* AC* at *P* . Prove that the tangent to the circle at *P* bisects *BC*.

#### Answer:

Given: ΔABC is right triangle in which ∠ABC = 90°. A circle is drawn with side AB as diameter intersecting AC in P.

PQ is the tangent to the circle when intersects BC in Q.

Construction: Join BP.

Proof:

PQ and BQ are tangents drawn from an external point Q.

⇒ PQ = BQ .....(i) [Length of tangents drawn from an external point to the circle are equal]

⇒ ∠PBQ = ∠BPQ [In a triangle, equal sides have equal angles opposite to them]

As , it is given that,

AB is the diameter of the circle.

∴ ∠APB = 90° [Angle in a semi-circle is 90°]

∠APB + ∠BPC = 180° [Linear pair]

⇒ ∠BPC = 180° – ∠APB = 180° – 90° = 90°

In ΔBPC,

∠BPC + ∠PBC + ∠PCB = 180° [Angle sum property]

⇒ ∠PBC + ∠PCB = 180° – ∠BPC = 180° – 90° = 90° ....(ii)

Now,

∠BPC = 90°

⇒ ∠BPQ + ∠CPQ = 90° .....(iii)

From (ii) and (iii), we get,

⇒ ∠PBC + ∠PCB = ∠BPQ + ∠CPQ

⇒ ∠PCQ = ∠CPQ [∠BPQ = ∠PBQ]

In ΔPQC,

∠PCQ = ∠CPQ

∴ PQ = QC .....(iv)

From (i) and (iv), we get,

BQ = QC

Thus, tangent at P bisects the side BC.

#### Page No 8.34:

#### Question 14:

From an external point *P*, tangents *PA* and *PB* are drawn to a circle with centre *O*. If *CD* is the tangent to the circle at a point *E* and *PA* = 14 cm, find the perimeter of Δ *PCD*.

#### Answer:

Let us first put the given data in the form of a diagram.

It is given that PA = 14cm. we have to find the perimeter of.

Perimeter of is *PC + CD + PD*

Looking at the figure we can rewrite the equation as follows.

Perimeter of is *PC + CE + ED + PD* ……(1)

From the property of tangents we know that the length of two tangents drawn to a circle from the same external point will be equal. Therefore,

*CE =CA*

*ED =DB*

Replacing the above in equation (1), we have,

Perimeter of as *PC + CA + DB + PD*

By looking at the figure we get,

*PC +CA =PA*

*DB +PD =PB*

Therefore,

Perimeter of is *PA + PB*

It is given that PA = 14 cm. again from the same property of tangents which says that the length of two tangents drawn to a circle from the same external point will be equal, we have,

*PA = PB*

Therefore,

Perimeter of = 2*PA*

Perimeter of = 2 × 14

Perimeter of = 28

Thus perimeter of is 28 cm.

#### Page No 8.34:

#### Question 15:

In the given figure, ABC is a right triangle right-angled at B such that BC = 6 cm and AB = 8 cm. Find the radius of its incircle.

#### Answer:

From the property of tangents we know that the length of two tangents drawn to a circle from the same external point will be equal. Therefore, we have

*BQ = BP*

Let us denote *BP* and *BQ* by *x*

*AP = AR*

Let us denote *AP* and *AR* by *y*

*RC = QC*

Let us denote *RC* and *RQ* by *z*

We have been given that is a right triangle and *BC* = 6 cm and *AB* = 8 cm. let us find out *AC* using Pythagoras theorem. We have,

Consider the perimeter of the given triangle. We have,

*AB + BC + AC* = 8 + 6 + 10

*AB + BC + AC* = 24

Looking at the figure, we can rewrite it as,

*AP + PB + BQ + QC + AR + RC* = 24

Let us replace the sides with the respective *x, y* and *z* which we have decided to use.

Now, consider the side *AC* of the triangle.

*AC = 10*

Looking at the figure we can say,

*AR + RC = 10*

*y + z = 10* …… (2)

Now let us subtract equation (2) from equation (1). We have,

*x + y + z* = 12

*y + z* = 10

After subtracting we get,

*x* = 2

That is,

*BQ* = 2, and

*BP* = 2

Now consider the quadrilateral *BPOQ*. We have,

*BP = BQ* (since length of two tangents drawn to a circle from the same external point are equal)

Also,

*PO = OQ* (radii of the same circle)

It is given that .

From the property of tangents, we know that the tangent will be at right angle to the radius of the circle at the point of contact. Therefore,

We know that sum of all angles of a quadrilateral will be equal to . Therefore,

Since all the angles of the quadrilateral are equal to and the adjacent sides also equal, this quadrilateral is a square. Therefore, all sides will be equal. We have found out that,

*BP* = 2 cm

Therefore, the radii

*PO* = 2 cm

Thus the radius of the incircle of the triangle is 2 cm.

#### Page No 8.35:

#### Question 16:

Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

#### Answer:

In the given figure, C is the mid point of the minor arc AB of the circle with centre O.

PQ is the tangent to the given circle through point C.

To Prove: Tangent drawn at the mid point of the arc $\stackrel{\u23dc}{AB}$ of a circle is parallel to the chord joining the end point of the arc $\stackrel{\u23dc}{AB}$.

i.e AB || PQ.

Proof: C is the mid point of the minor arc AB

⇒ minor arc AC = minor arc BC

⇒ AC = BC

Thus, ∆ABC is an iscosceles triangle.

Therefore, the perpendicular bisector of side AB of ∆ABC passes through the vertex C.

We know that the perpendicular bisector of a chord passes through the centre of the circle.

Since AB is a chord of the circle so, the perpendicular bisector of AB passes through the centre O.

Thus, it is clear that the perpendicular bisector of AB passes through the points O and C.

Therefore, $\mathrm{AB}\perp \mathrm{OC}$

Now, PQ is the tangent to the circle through the point C on the circle.

Therefore, $\mathrm{PQ}\perp \mathrm{OC}\left[\mathrm{Tangent}\mathrm{to}\mathrm{a}\mathrm{circle}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{its}\mathrm{radius}\mathrm{through}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{contact}\right]\phantom{\rule{0ex}{0ex}}$

#### Page No 8.35:

#### Question 17:

From a point *P*, two tangents *PA* and *PB* are drawn to a circle with centre *O*. If *OP* = diameter of the circle, show that Δ *APB *is equilateral.

#### Answer:

Let us first put the given data in the form of a diagram.

Consider and. We have,

*PO* is the common side for both the triangles.

*PA = PB*(Tangents drawn from an external point will be equal in length)

*OB = OA*(Radii of the same circle)

Therefore, by SSS postulate of congruency, we have

ΔPOA ΔPOB

Hence,

…… (1)

Now let us consider and . We have,

*PL* is the common side for both the triangles.

(From equation (1))

*PA = PB* (Tangents drawn from an external point will be equal in length)

From SAS postulate of congruent triangles,

Therefore,

*PL = LB *…… (2)

Since AB is a straight line,

Let us now take up ΔOPB. We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore,

By Pythagoras theorem we have,

It is given that

*OP* = diameter of the circle

Therefore,

*OP = 2OB*

Hence,

Consider . We have,

But we have found that,

Also from the figure, we can say

*PL = PO − OL *

Therefore,

…… (3)

Also, from , we have

…… (4)

Since Left Hand Sides of equation (3) and equation (4) are same, we can equate the Right Hand Sides of the two equations. Thus we have,

=

We know from the given data, that OP = 2.OB. Let us substitute 2OB in place of PO in the above equation. We get,

Substituting the value of *OL* and also *PO* in equation (3), we get,

Also from the figure we get,

*AB = PL + LB*

From equation (2), we know that *PL* = *LB*. Therefore,

*AB = 2.LB*

We have also found that. We know that tangents drawn from an external point will be equal in length. Therefore, we have

*PA = PB*

Hence,

*PA* =

Now, consider . We have,

*PA* =

*PB* =

*AB* =

Since all the sides of the triangle are of equal length, is an equilateral triangle. Thus we have proved.

#### Page No 8.35:

#### Question 18:

Two tangent segments *PA* and PB are drawn to a circle with centre *O* such that ∠*APB* = 120°. Prove that OP = 2 *AP*.

#### Answer:

Let us first put the given data in the form of a diagram. We have,

Consider and . We have,

Here, PO is the common side.

PA = PB (Length of two tangents drawn from the same external point will be equal)

OA = OB(Radii of the same circle)

By SSS congruency, we have is congruent to .

Therefore,

It is given that,

That is,

(Since )

In ,

(Since radius will be perpendicular to the tangent at the point of contact)

We know that,

Thus we have proved.

#### Page No 8.35:

#### Question 19:

If Δ *ABC* is isosceles with *AB* = *AC* and *C* (*O,* *r*) is the incircle of the Δ*ABC* touching *BC* at *L*,prove that *L* bisects *BC*.

#### Answer:

Let us first put the given data in the form of a diagram.

It is given that triangle *ABC* is isosceles with

*AB = AC …*… (1)

By looking at the figure we can rewrite the above equation as,

*AM + MB = AN + NC*

From the property of tangents we know that the length of two tangents drawn to a circle from the same external point will be equal. Therefore,

*AM = AN*

Let us substitute *AN* with *AM* in the equation (1). We get,

*AM + MB = AM + NC*

*MB = NC* …… (2)

From the property of tangents we know that the length of two tangents drawn from the same external point will be equal. Therefore we have,

*MB = BL*

*NC = LC*

But from equation (2), we have found that

*MB = NC*

Therefore,

*BL = LC*

Thus we have proved that point L bisects side *BC*.

#### Page No 8.35:

#### Question 20:

*AB* is a diameter and* *AC is a chord of a circle with centre *O *such that $\angle BAC={30}^{\xb0}$ . The tangent at* C* intersects *AB* at a point *D* . Prove that *BC* =* BD*.

#### Answer:

It is given that ∠BAC = 30° and AB is diameter.

$\frac{\mathrm{AB}}{2}=\mathrm{OA}=\mathrm{OC}\left(\mathrm{Radius}\right)$

∠ACB = 90° (Angle formed by the diameter is 90°)

In ∆ABC,

∠ACB + ∠BAC + ∠ABC = 180°

⇒ 90° + 30° + ∠ABC = 180°

⇒ ∠ABC = 60°

⇒ ∠CBD = 180° – 60° = 120° ( ∠CBD and ∠ABC form a linear pair)

In ∆OCD,

∠OCD = 90° (Angle made by Radius on the tangent)

∠OBC = ∠ABC = 60°

Since OB = OC, ∠OCB = ∠OBC = 60° (OC = OB = radius)

In ∆OCB,

⇒ ∠COB + ∠OCB + ∠OBC = 180°

⇒ ∠COB + 60° + 60° = 180°

⇒ ∠COB = 60°

In ∆OCD,

∠COD + ∠OCD + ∠ODC = 180°

⇒ 60° + 90° + ∠ODC = 90° (∠COD = ∠COB)

⇒ ∠ODC = 30°

In ∆CBD,

∠CBD = 120°

∠BDC = ∠ODC = 30°

⇒ ∠BCD + ∠BDC + ∠CBD = 180°

⇒ ∠BCD + 30° + 120° = 180°

⇒ ∠BCD + 30° = ∠BDC

Angles made by BC and BD on CD are equal, so ∆CBD is an isosceles triangle and therefore, BC = BD.

#### Page No 8.35:

#### Question 21:

In the given figure, a circle touches all the four sides of a quadrilateral *ABCD* with *AB* = 6 cm, *BC* = 7 cm and *CD *= 4 cm. Find *AD*.

#### Answer:

The figure given in the question is below.

From the property of tangents we know that, the length of two tangents drawn from the same external point will be equal. Therefore we have the following,

*SA = AP*

For our convenience, let us represent *SA* and *AP* by *a*

*PB = BQ*

Let us represent *PB* and *BQ* by *b*

*QC = CR*

Let us represent *QC* and *CR* by *c*

*DR = DS*

Let us represent *DR* and *DS* by *d*

It is given in the problem that,

*AB* = 6

By looking at the figure we can rewrite the above equation as follows,

*AP + PB* = 6

*a + b* = 6

*b* = 6 − *a* …… (1)

Similarly we have,

*BC* = 7

*BQ + QC* = 7

*b + c* = 7

Let us substitute the value of b which we have found in equation (1). We have,

6 − *a + c* = 7

*c = a* + 1 …… (2)

*CD* = 4

*CR + RD* = 4

*c + d* = 4

Let us substitute the value of c which we have found in equation (2).

*a* + 1 + *d* = 4

*a + d* =3

As per our representations in the previous section, we can write the above equation as follows,

*SA + DS* = 3

By looking at the figure we have,

*AD* = 3

Thus we have found that length of side *AD* is 3 cm.

#### Page No 8.35:

#### Question 22:

Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.

#### Answer:

Let us first put the given data in the form of a diagram. Let us also draw another parallel tangent *PQ* parallel to the given tangent *AB*. Draw a radius *OR* to the point of contact of the tangent *PQ*. Let us also draw a radius *CO* parallel to the two tangents.

It is given that,

We know that sum of angles on the same side of the transversal will be equal to 180°. Therefore,

We know that the radius of the circle will always be perpendicular to the tangent at the point of contact. Therefore,

Again, since sum of angles on the same side of the transversal is equal to 180°, we have,

Now let us add equation (1) and (2). We get,

Looking at the figure we can rewrite the above equation as,

We know that the angle of a straight line will measure 180°. Therefore, *ROS* is the straight line. Also, *RO* is the radius which we have drawn and we know that a radius is always drawn from the centre of the circle. Therefore, line *ROS* passes through the centre of the circle.

Thus we have proved that the perpendicular to the tangent at the point of contact passes through the centre of the circle.

#### Page No 8.35:

#### Question 23:

Two circles touch externally at a point P. From a point T on the tangent at P, tangents TQ and TR are drawn to the circles with points of contact Q and R respectively. Prove that TQ = TR.

#### Answer:

We know that the lengths of tangents drawn from an external point to a circle are equal.

In the given figure, TQ and TP are tangents drawn to the same circle from an external point T.

∴ TQ = TP .....(1)

Also, TP and TR are tangents drawn to the same circle from an external point T.

∴ TP = TR .....(2)

From (1) and (2), we get

TQ = TR

#### Page No 8.36:

#### Question 24:

*A *is a point at a distance 13 cm from the centre* O* of a circle of radius 5 cm . *AP* and *AQ *are the tangents to the circle at *P* and *Q* . If a tangent* BC* is drawn at a point *R* lying on the minor arc *PQ* to intersect *AP* at *B * and *AQ* at *C* , find the perimeter of the $\u2206$* ABC* .

#### Answer:

A is a point 13 cm from centre O. AP and AQ are the tangents to the circle with centre O.

AP = AQ

In ∆APO,

${\mathrm{AP}}^{2}+{\mathrm{OP}}^{2}={\mathrm{AO}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AP}}^{2}={\mathrm{AO}}^{2}-{\mathrm{OP}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AP}}^{2}={13}^{2}-{5}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AP}}^{2}=169-25=144\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AP}=12\mathrm{cm}$

Now In ∆ABC,

Perimeter = AB + BC + AC

= AB + BR + RC + AC

= AB + BP + CQ + AC ( BR = BP, RC = CQ)

= AP + AQ

= 12 + 12

= 24 cm

#### Page No 8.36:

#### Question 25:

In the given figure, a circle is inscribed in a quadrilateral *ABCD* in which ∠*B** *= 90°. It AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius *r *of the circle.

#### Answer:

Let us first consider the quadrilateral *OPBQ*.

It is given that .

Also from the property of tangents we know that the radius of the circle will always be perpendicular to the tangent at the point of contact. Therefore we have,

We know that sum of all angles of a quadrilateral will always be equal to . Therefore,

Also, in the quadrilateral,

*OQ = OP* (both are the radii of the same circle)

*PB = BQ* (from the property of tangents which says that the length of two tangents drawn to a circle from the same external point will be equal)

Since the adjacent sides of the quadrilateral are equal and also since all angles of the quadrilateral are equal to 90°, we can conclude that the quadrilateral OPBQ is a square.

It is given that *DS* = 5 cm.

From the property of tangents we know that the length of two tangents drawn from the same external point will be equal. Therefore,

*DS = DR*

*DR* = 5

It is given that,

*AD* = 23

*DR + RA *= 23

5 + *RA* = 23

*RA* = 18

Again from the same property of tangents we have,

*RA = AQ*

We have found out *RA* = 18. Therefore,

*AQ* = 18

It is given that *AB* = 29. That is,

*AQ* + *QB* = 29

18 + *QB* = 29

*QB* = 11

We have initially proved that *OPBQ* is a square. *QB* is one of the sides of the square. Since all sides of the square will be of equal length, we have,

*OP = 11*

*OP* is nothing but the radius of the circle.

Thus the radius of the circle is equal to 11 cm.

#### Page No 8.36:

#### Question 26:

In the given figure, there are two concentric circles with centre *O* of radii 5 cm and 3 cm. From an external point *P*, tangent *PA* and *PB* are drawn to these circles. If *AP* = 12 cm, find the length of *BP*.

#### Answer:

The figure given the question is

From the property of tangents we know that the radius of the circle will always be perpendicular to the tangent at the point of contact. Therefore, *OA* is perpendicular to *AP* and triangle *OAP* is a right triangle. Therefore,

Now consider the smaller circle. Here again the radius OB will be perpendicular to the tangent BP. Therefore, triangle OBP is a right triangle. Hence we have,

Thus we have found that the length of BP is .

#### Page No 8.36:

#### Question 27:

In the given figure, *AB* is a chord of length 16 cm of a circle of radius 10 cm. The tangents at *A* and* B* intersect at a point *P*. Find the length of *PA*.

#### Answer:

Consider and .

From the property of tangents we know that the length of two tangents drawn form an external point will be equal. Therefore we have,

*PA = PB*

*OB = OA* (They are the radii of the same circle)

*PO* is the common side

Therefore, from SSS postulate of congruency, we have,

Hence,

…… (1)

Now consider and . We have,

(From (1))

*PA* is the common side.

From the property of tangents we know that the length of two tangents drawn form an external point will be equal. Therefore we have,

*PA = PB*

From SAS postulate of congruent triangles, we have,

Therefore,

*LA = LB*

It is given that AB = 16. That is,

*LA + LB* = 16

*LA + LA* = 16

*2LA* = 16

*LA* = 8

*LB* = 8

Also, *ALB* is a straight line. Therefore,

That is,

Since ,

Therefore,

Now let us consider. We have,

Consider. Here,

(Since the radius of the circle will always be perpendicular to the tangent at the point of contact)

Therefore,

…… (1)

Now consider

…… (2)

Since the Left Hand Side of equation (1) is same as the Left Hand Side of equation (2), we can equate the Right Hand Side of the two equations. Hence we have,

= …… (3)

From the figure we can see that,

*OP = OL + LP*

Therefore, let us replace *OP* with *OL + LP* in equation (3). We have,

We have found that *OL* = 6 and *LB* = 8. Also it is given that *OB* = 10. Substituting all these values in the above equation, we get,

Now, let us substitute the value of *PL* in equation (2). We get,

We know that tangents drawn from an external point will always be equal. Therefore,

*PB = PA*

Hence, we have,

*PA* =

#### Page No 8.37:

#### Question 28:

In the given figure, *PA *and *PB *are tangents from an external point *P* to a circle with centre *O*. *LN *touches the circle at M. Prove that *PL* + *LM* = *PN* + *MN*.

#### Answer:

The figure given in the question

From the property of tangents we know that the length of two tangents drawn from an external point will we be equal. Hence we have,

*PA = PB*

*PL + LA = PN + NB* …… (1)

Again from the same property of tangents we have,

*LA = LM* (where L is the common external point for tangents LA and LM)

*NB = MN* (where N is the common external point for tangents NB and MN)

Substituting *LM* and *MN* in place of *LA* and *NB* in equation (1), we have

*PL + LM = PN + MN*

Thus we have proved.

#### Page No 8.37:

#### Question 29:

In the given figure, *BDC* is a tangent to the given circle at point *D* such that *BD* = 30 cm and *CD *= 7 cm. The other tangents *BE* and *CF* are drawn respectively from *B* and *C* to the circle and meet when produced at *A* making BAC a right angle triangle. Calculate (i) *AF* (ii) radius of the circle.

#### Answer:

The given figure is below

(i) The given triangle *ABC* is a right triangle where side BC is the hypotenuse. Let us now apply Pythagoras theorem. We have,

Looking at the figure we can rewrite the above equation as follows.

…… (1)

From the property of tangents we know that the length of two tangents drawn from the same external point will be equal. Therefore we have the following,

BE = BD

It is given that BD = 30 cm. Therefore,

BE = 30 cm

Similarly,

CD = FC

It is given that CD = 7 cm. Therefore,

FC = 7 cm

Also, on the same lines,

EA = AF

Let us substitute these in equation (1). We get,

Therefore,

*AF* = 5

Or,

*AF* = − 42

Since length cannot have a negative value,

*AF* = 5

(ii) Let us join the point of contact *E* with the centre of the circle say *O*. Also, let us join the point of contact *F* with the centre of the circle *O*. Now we have a quadrilateral *AEOF*.

e

In this quadrilateral we have,

(Given in the problem)

(Since the radius will always be perpendicular to the tangent at the point of contact)

(Since the radius will always be perpendicular to the tangent at the point of contact)

We know that the sum of all angles of a quadrilateral will be equal to . Therefore,

Since all the angles of the quadrilateral are equal to 90° and the adjacent sides are equal, this quadrilateral is a square. Therefore all the sides are equal. We have found that

*AF* = 5

Therefore,

*OD* = 5

*OD* is nothing but the radius of the circle.

Thus we have found that *AF* = 5 cm and radius of the circle is 5 cm.

#### Page No 8.37:

#### Question 30:

If ${d}_{1},{d}_{2}({d}_{2}{d}_{1})$ be the diameters of two concentric circle s and *c* be the length of a chord of a circle which is tangent to the other circle , prove that ${{d}_{2}}^{2}={c}^{2}+{{d}_{1}}^{2}$.

#### Answer:

Let O be the centre of two concentric circles and PQ be the tangent to the inner circle that touches the circle at R.

Now, OQ= $\frac{1}{2}{d}_{2}$ and OR= $\frac{1}{2}{d}_{1}$

Also, PQ = *c*

As, PQ is the tangent to the circle.

⇒ OR ⊥ PQ

⇒ QR =$\frac{1}{2}\mathrm{PQ}=\frac{1}{2}c$

In Triangle OQR,

∴ By Pythagoras Theorem,

${\left(OQ\right)}^{2}={\left(OR\right)}^{2}+{\left(RQ\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{{d}_{2}}{2}\right)}^{2}={\left(\frac{{d}_{1}}{2}\right)}^{2}+{\left(\frac{c}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left({d}_{2}\right)}^{2}={\left({d}_{1}\right)}^{2}+{c}^{2}$

#### Page No 8.37:

#### Question 31:

In the given figure, tangents PQ and PR are drawn from an external point to a circle with centre O, such that $\angle \mathrm{RPQ}=30\xb0$. A chord RS is drawn parallel to the tangent PQ. Find $\angle \mathrm{RQS}$. [CBSE 2015]

#### Answer:

It is given that tangents PQ and PR are drawn from an external point P to a circle with centre O, such that $\angle \mathrm{RPQ}=30\xb0$. Also, RS || PQ.

Join OR and OS.

PQ and PR are tangents drawn from an external point P to a circle.

∴ PQ = PR (Lengths of tangents drawn from an external point to a circle are equal)

In ∆PQR,

PQ = PR

∴ $\angle \mathrm{PRQ}=\angle \mathrm{PQR}$ (In a triangle, equal sides have equal angles opposite to them)

Now,

$\angle \mathrm{PRQ}+\angle \mathrm{PQR}+\angle \mathrm{RPQ}=180\xb0$ (Angle sum property)

$\Rightarrow 2\angle \mathrm{PRQ}+30\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 2\angle \mathrm{PRQ}=180\xb0-30\xb0=150\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{PRQ}=75\xb0\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{PQR}=\angle \mathrm{PRQ}=75\xb0.....\left(1\right)$

Since SR || PQ and RQ is the transversal,

∴ $\angle \mathrm{SRQ}=\angle \mathrm{RQP}=75\xb0$ (Alternate angles)

Now, PR is the tangent and OR is the radius through the point of contact R.

∴ $\angle \mathrm{ORP}=90\xb0$ (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

$\Rightarrow \angle \mathrm{ORQ}=\angle \mathrm{ORP}-\angle \mathrm{PRQ}=90\xb0-75\xb0=15\xb0\left[\mathrm{Using}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{ORS}=\angle \mathrm{SRQ}-\angle \mathrm{ORQ}=75\xb0-15\xb0=60\xb0$

In ∆SOR,

OR = OS (Radii of the circle)

∴ $\angle \mathrm{OSR}=\angle \mathrm{ORS}=60\xb0$ (In a triangle, equal sides have equal angles opposite to them)

Now,

$\angle \mathrm{ORS}+\angle \mathrm{OSR}+\angle \mathrm{ROS}=180\xb0$ (Angle sum property)

$\Rightarrow 60\xb0+60\xb0+\angle \mathrm{ROS}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{ROS}=180\xb0-120\xb0=60\xb0$

We know that the angle subtended by an arc at the centre is twice the angle subtended by it any point on the remaining part of the circle.

$\therefore \angle \mathrm{ROS}=2\angle \mathrm{RQS}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{RQS}=\frac{1}{2}\angle \mathrm{ROS}=\frac{1}{2}\times 60\xb0=30\xb0$

#### Page No 8.37:

#### Question 32:

From an external point *P* , tangents *PA *= *PB *are drawn to a circle with centre *O* . If $\angle PAB={50}^{\xb0}$, then find $\angle AOB$.

#### Answer:

It is given that PA and PB are tangents to the given circle.

$\therefore \angle \mathrm{PAO}=90\xb0$ (Radius is perpendicular to the tangent at the point of contact.)

Now,

$\angle \mathrm{PAB}=50\xb0$ (Given)

$\therefore \angle \mathrm{OAB}=\angle \mathrm{PAO}-\angle \mathrm{PAB}=90\xb0-50\xb0=40\xb0$

In ∆OAB,

OB = OA (Radii of the circle)

$\therefore \angle \mathrm{OAB}=\angle \mathrm{OBA}=40\xb0$ (Angles opposite to equal sides are equal.)

Now,

$\angle \mathrm{AOB}+\angle \mathrm{OAB}+\angle \mathrm{OBA}=180\xb0$ (Angle sum property)

$\Rightarrow \angle \mathrm{AOB}=180\xb0-40\xb0-40\xb0=100\xb0$

#### Page No 8.38:

#### Question 33:

In the given figure, two tangents *AB* and *AC* are drawn to a circle with centre *O* such that ∠*BAC* = 120°. Prove that *OA* = 2*AB*.

#### Answer:

Consider and.

We have,

*OB = OC* (Since they are radii of the same circle)

*AB = AC* (Since length of two tangents drawn from an external point will be equal)

*OA* is the common side.

Therefore by SSS congruency, we can say that andare congruent triangles.

Therefore,

It is given that,

We know that,

We know that,

Therefore,

=

*OA = 2AB*

#### Page No 8.38:

#### Question 34:

The length of three concesutive sides of a quadrilateral circumscribing a circle are 4 cm, 5 cm, and 7 cm respectively. Determine the length of the fourth side.

#### Answer:

Let us first put the given data in the form of a diagram.

From the property of tangents we know that the length of two tangents drawn from the same external point will be equal. Therefore we have,

*AR = SA*

Let us represent *AR* and *SA* by ‘*a’*.

Similarly,

*QB = RB*

Let us represent SD and DP by *‘b’*

*PC = CQ*

Let us represent *PC* and *PQ* by ‘*c’*

*SD = DP*

Let us represent *QB* and *RB* by ‘*d*’

It is given that,

*AB* = 4

*AR + RB* =4

*a + b* = 4

*b* = 4 − *a* …… (1)

Similarly,

*BC* = 5

That is,

*b + c* = 5

Let us substitute for *b* from equation (1). We get,

4 − *a + c* = 5

*c − a* = 1

*c = a* + 1 …… (2)

*CD* = 7

*c + d* = 7

Let us substitute for *c* from equation (2). We get,

*a* + 1 + *d* = 7

*a + d *= 6

In the previous section we had represented AS and SR with ‘*a*’ and SD and DP with ‘*b*’. We shall now put AS in place of ‘*a*’ and SD in place of ‘*d*’. We get,

*AS + SD* = 6

*AD* = 6 cm

Therefore, the length of the fourth side of the quadrilateral is 6 cm.

#### Page No 8.38:

#### Question 35:

The common tangents *AB* and *CD* to two circles with centres *O* and *O'* intersect at *E *between their centres . Prove that the points *O *, *E *and *O'* are collinear .

#### Answer:

Here Angle AEC and DEB are equal ( vertically opposite angles)

Join OA and OC,

So in triangle OAE and OCE, we have

OA = OC ( radii of same circle)

OE = OE (common)

$\angle $OAE = $\angle $OCE [90 each, as tangent is always perpendicular to its radius at point of contact]

$\u25b3OEA\cong \u25b3OCE\left(\mathrm{RHS}\right)\phantom{\rule{0ex}{0ex}}So,\angle AEO=\angle CEO\left(\mathrm{CPCT}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Similarly},\mathrm{for}\mathrm{the}\mathrm{other}\mathrm{circle}\mathrm{we}\mathrm{have}\phantom{\rule{0ex}{0ex}}\angle DEO\text{'}=\angle BEO\text{'}\phantom{\rule{0ex}{0ex}}\mathrm{Now}\angle AEC=\angle DEB\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\angle AEC=\frac{1}{2}\angle DEB\phantom{\rule{0ex}{0ex}}\Rightarrow \angle AEO=\angle CEO=\angle DEO\text{'}=\angle BEO\text{'}$

So, all four angles are equal and bisected by OE and OE'.

Hence, O, E' and O' are collinear.

#### Page No 8.38:

#### Question 36:

In the given figure, common tangents *PQ* and *RS* to two circles intersect at *A*. Prove that *PQ* = *RS.*

#### Answer:

The figure given in the question is

We know from the property of tangents that the length of two tangents drawn from a common external point will be equal. Therefore,

*PA = RA* …… (1)

*AQ = AS* …… (2)

Let us add equation (1) and (2)

*PA + AQ = RA + AS*

*PQ = RS*

Thus we have proved that *PQ = RS.*

#### Page No 8.38:

#### Question 37:

Two concentric circles are of diameters 30 cm and 18 cm. Find the length of the chord of the larger circle which touches the smaller circle. [CBSE 2014]

#### Answer:

Let O be the common centre of the two circles and AB be the chord of the larger circle which touches the smaller circle at C.

Join OA and OC. Then,

OC = $\frac{18}{2}$ cm = 9 cm and OA = $\frac{30}{2}$ cm = 15 cm

We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Also, the perpendicular drawn from the centre of a circle to a chord bisects the chord.

∴ OC ⊥ AB and C is the mid-point of AB.

In right ∆OCA,

${\mathrm{OA}}^{2}={\mathrm{OC}}^{2}+{\mathrm{AC}}^{2}\left(\mathrm{Pythagoras}\mathrm{theorem}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}={\mathrm{OA}}^{2}-{\mathrm{OC}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=\sqrt{{15}^{2}-{9}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=\sqrt{225-81}=\sqrt{144}=12\mathrm{cm}$

∴ AB = 2AC = 2 × 12 cm = 24 cm

Thus, the required length of the chord is 24 cm.

#### Page No 8.38:

#### Question 38:

*AB* and *CD* are common tangents to two circles of equal radii. Prove that *AB* = *CD*.

#### Answer:

*Given: Two circles with centre’s O** and O'**. AB and CD are common tangents to the circles which intersect in P.*

*To Prove: AB = CD*

*Proof:*

*AP = PC **(length of tangents drawn from an external point to the circle are equal)** ..… (1) *

*PB = PD **(length of tangents drawn from an external point to the circle are equal)** ..… (2) *

*Adding (1) and (2), we get*

*AP + PB = PC + PD*

*⇒ **AB = CD
Hence Proved*

#### Page No 8.38:

#### Question 39:

A triangle PQR is drawn to circumscribe a circle of radius 8 cm such that the segments QT and TR, into which QR is divided by the point of contact T, are of lengths 14 cm and 16 cm respectively. If area of ∆PQR is 336 cm^{2}, find the sides PQ and PR. [CBSE 2014]

#### Answer:

Here, T, S and U are the points of contact of the circle with the sides QR, PQ and PR, respectively.

OT = OS = OU = 8 cm (Radii of the circle)

We know that the lengths of tangents drawn from an external point to a circle are equal.

∴ QS = QT = 14 cm

RU = RT = 16 cm

PS = PU = *x* cm (say)

So, QR = QT + TR = 14 cm + 16 cm = 30 cm

PQ = PS + SQ = *x* cm + 14 cm = (*x* + 14) cm

PR = PU + UR = *x* cm + 16 cm = (*x* + 16) cm

Also, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ OT ⊥ QR, OS ⊥ PQ and OU ⊥ PR

Now,

ar(∆OQR) + ar(∆OPQ) + ar(∆OPR) = ar(∆PQR)

$\therefore \frac{1}{2}\times \mathrm{QR}\times \mathrm{OT}+\frac{1}{2}\times \mathrm{PQ}\times \mathrm{OS}+\frac{1}{2}\times \mathrm{PR}\times \mathrm{OU}=336{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\times 30\times 8+\frac{1}{2}\times \left(x+14\right)\times 8+\frac{1}{2}\times \left(x+16\right)\times 8=336\phantom{\rule{0ex}{0ex}}\Rightarrow 120+4x+56+4x+64=336\phantom{\rule{0ex}{0ex}}\Rightarrow 8x+240=336$

$\Rightarrow 8x=336-240=96\phantom{\rule{0ex}{0ex}}\Rightarrow x=12$

∴ PQ = (*x* + 14) cm = (12 + 14) cm = 26 cm

PR = (*x* + 16) cm = (12 + 16) cm = 28 cm

Hence, the lengths of sides PQ and PR are 26 cm and 28 cm, respectively.

#### Page No 8.39:

#### Question 40:

In Fig . 10.69, the tangent at a point *C *of a circle and a diameter *AB* when extended intersect at *P* . If $\angle $*PCA* =110^{0}, find $\angle $*CBA*. [**Hint**: Join *CO*.]

figure

#### Answer:

$\angle \mathrm{ACB}=90\xb0\left(\mathrm{Angle}\mathrm{inscribed}\mathrm{in}\mathrm{a}\mathrm{semi}-\mathrm{circle}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{PCO}=90\xb0\left(\mathrm{PC}\mathrm{is}\mathrm{a}\mathrm{tangent}\mathrm{at}\mathrm{C}\right)\phantom{\rule{0ex}{0ex}}$

$Now,\angle \mathrm{PCA}=\angle \mathrm{PCO}+\angle \mathrm{OCA}\phantom{\rule{0ex}{0ex}}\Rightarrow 110\xb0=90\xb0+\angle \mathrm{OCA}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{OCA}=20\xb0$

Since, OC = OA (radii of the circle)

$\angle \mathrm{OCA}=\angle \mathrm{OAC}=20\xb0\phantom{\rule{0ex}{0ex}}In\u2206ABC,\phantom{\rule{0ex}{0ex}}\angle \mathrm{BAC}+\angle \mathrm{ACB}+\angle \mathrm{CBA}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 90\xb0+20\xb0+\angle \mathrm{CBA}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{CBA}=70\xb0$

#### Page No 8.39:

#### Question 41:

*AB* is a chord of a circle with centre *O* , *AOC* is a diameter and *AT* is the tangent at* A* as shown in Fig . 10.70. Prove that $\angle $*BAT* = $\angle $*ACB*.

#### Answer:

In the given figure,

AC is the diameter.

So, $\angle CBA=90\xb0$ (Angle formed by the diameter on the circle is 90º)

AT is the tangent at point A.

Thus, $\angle \mathrm{CAT}=90\xb0$

In ∆ABC,

$\angle \mathrm{BCA}+\angle \mathrm{BAC}+\angle \mathrm{CBA}=180\xb0\left(\mathrm{Angle}\mathrm{sum}\mathrm{property}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{BCA}+90\xb0+\angle CAT-\angle BAT=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{BCA}+90\xb0+90-\angle BAT=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{BCA}=\angle BAT$

Hence Proved

#### Page No 8.39:

#### Question 42:

In the given figure, a ∆ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 8 cm and 6 cm respectively. Find the lengths of sides AB and AC, when area of ∆ABC is 84 cm^{2}. [CBSE 2015]

#### Answer:

Here, D, E and F are the points of contact of the circle with the sides BC, AB and AC, respectively.

OD = OE = OF = 4 cm (Radii of the circle)

We know that the lengths of tangents drawn from an external point to a circle are equal.

∴ BD = BE = 8 cm

CD = CF = 6 cm

AE = AF = *x* cm (say)

So, BC = BD + CD = 8 cm + 6 cm = 14 cm

AB = AE + BE = *x* cm + 8 cm = (*x* + 8) cm

AC = AF + FC = *x* cm + 6 cm = (*x* + 6) cm

Also, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ OD ⊥ BC, OE ⊥ AB and OF ⊥ AC

Now,

ar(∆OBC) + ar(∆OAB) + ar(∆OCA) = ar(∆ABC)

$\therefore \frac{1}{2}\times \mathrm{BC}\times \mathrm{OD}+\frac{1}{2}\times \mathrm{AB}\times \mathrm{OE}+\frac{1}{2}\times \mathrm{AC}\times \mathrm{OF}=84{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\times 14\times 4+\frac{1}{2}\times \left(x+8\right)\times 4+\frac{1}{2}\times \left(x+6\right)\times 4=84\phantom{\rule{0ex}{0ex}}\Rightarrow 28+2x+16+2x+12=84\phantom{\rule{0ex}{0ex}}\Rightarrow 4x+56=84$

$\Rightarrow 4x=84-56=28\phantom{\rule{0ex}{0ex}}\Rightarrow x=7$

∴ AB = (*x* + 8) cm = (7 + 8) cm = 15 cm

AC = (*x* + 6) cm = (7 + 6) cm = 13 cm

Hence, the lengths of sides AB and AC are 15 cm and 13 cm, respectively.

#### Page No 8.39:

#### Question 43:

In the given figure, AB is a diameter of a circle with centre O and AT is a tangent. If $\angle $AOQ = 58º, find $\angle $ATQ. [CBSE 2015]

#### Answer:

It is given that $\angle $AOQ = 58º.

We know that the angle subtended by an arc at the centre is twice the angle subtended by it any point on the remaining part of the circle.

∴ $\angle \mathrm{ABQ}=\frac{1}{2}\angle \mathrm{AOQ}=\frac{1}{2}\times 58\xb0=29\xb0$

Now, AT is the tangent and OA is the radius of the circle through the point of contact A.

$\therefore \angle \mathrm{OAT}=90\xb0$ (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

In ∆ABT,

$\angle \mathrm{BAT}+\angle \mathrm{ABT}+\angle \mathrm{ATB}=180\xb0$ (Angle sum property)

$\Rightarrow 90\xb0+29\xb0+\angle \mathrm{ATB}=180\xb0\left(\angle \mathrm{BAT}=\angle \mathrm{OAT}\mathrm{and}\angle \mathrm{ABT}=\angle \mathrm{ABQ}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{ATB}=180\xb0-119\xb0=61\xb0\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{ATQ}=\angle \mathrm{ATB}=61\xb0$

#### Page No 8.40:

#### Question 44:

In the given figure, OQ : PQ = 3.4 and perimeter of Δ POQ = 60 cm. Determine PQ, QR and OP.

#### Answer:

In the figure,

. Therefore we can use Pythagoras theorem to find the side *PO*.

…… (1)

In the problem it is given that,

…… (2)

Substituting this in equation (1), we have,

…… (3)

It is given that the perimeter of is 60 cm. Therefore,

*PQ + OQ + PO* = 60

Substituting (2) and (3) in the above equation, we have,

Substituting for *PQ* in equation (2), we have,

*OQ* is the radius of the circle and *QR* is the diameter. Therefore,

*QR* = *2OQ*

*QR* = 30

Substituting for *PQ* in equation (3), we have,

Thus we have found that *PQ* = 20 cm, *QR* = 30 cm and *PO* = 25 cm.

#### Page No 8.40:

#### Question 45:

Equal circles with centres O and O' touch each other at X. OO' produced to meet a circle with centre O', at A. AC is a tangent to the circle whose centre is O. O'D is perpendicular to AC. Find the value of $\frac{DO\text{'}}{CO}$.

#### Answer:

Consider the two triangles and .

We have,

is a common angle for both the triangles.

(Given in the problem)

(Since *OC* is the radius and *AC* is the tangent to that circle at *C* and we know that the radius is always perpendicular to the tangent at the point of contact)

Therefore,

From *AA* similarity postulate we can say that,

~

Since the triangles are similar, all sides of one triangle will be in same proportion to the corresponding sides of the other triangle.

Consider *AO*′ of and *AO* of .

Since *AO*′ and *O*′*X* are the radii of the same circle, we have,

*AO*′* = O*′*X*

Also, since the two circles are equal, the radii of the two circles will be equal. Therefore,

*AO*′* = XO*

Therefore we have

Since ~,

We have found that,

Therefore,

#### Page No 8.40:

#### Question 46:

In the given figure, *BC* is a tangent to the circle with centre *O*. *OE* bisects *AP.* Prove that Δ*AEO* $~$Δ *ABC*.

#### Answer:

The figure given in the question is below

Let us first take up .

We have,

*OA = OP* (Since they are the radii of the same circle)

Therefore, is an isosceles triangle. From the property of isosceles triangle, we know that, when a median drawn to the unequal side of the triangle will be perpendicular to the unequal side. Therefore,

Now let us take up and .

We know that the radius of the circle will always be perpendicular to the tangent at the point of contact. In this problem, *OB* is the radius and *BC* is the tangent and *B* is the point of contact. Therefore,

Also, from the property of isosceles triangle we have found that

Therefore,

=

is the common angle to both the triangles.

Therefore, from AA postulate of similar triangles,

~

Thus we have proved.

#### Page No 8.40:

#### Question 47:

In the given figure, *PO *$\perp $ *QO*. The tangents to the circle at *P* and *Q* intersect at a point *T*. Prove that *PQ *and *OT* are right bisector of each other.

#### Answer:

In the given figure,

*PO = OQ (Since they are the radii of the same circle)*

*PT = TQ *(Length of the tangents from an external point to the circle will be equal) Now considering the angles of the quadrilateral PTQO, we have,

(Given in the problem)

(The radius of the circle will be perpendicular to the tangent at the point of contact)

(The radius of the circle will be perpendicular to the tangent at the point of contact)

We know that the sum of all angles of a quadrilateral will be equal to . Therefore,

Thus we have found that all angles of the quadrilateral are equal to 90°.

Since all angles of the quadrilateral *PTQO* are equal to 90° and the adjacent sides are equal, this quadrilateral is a square.

We know that in a square, the diagonals will bisect each other at right angles.

Therefore, *PQ* and *OT* bisect each other at right angles.

Thus we have proved.

#### Page No 8.41:

#### Question 48:

In the given figure, O is the centre of the circle and *BCD *is tangent to it at *C*. Prove that ∠*BAC* + ∠*ACD* = 90°.

#### Answer:

In the given figure, let us join *D* an *A*.

Consider . We have,

*OC = OA* (Radii of the same circle)

We know that angles opposite to equal sides of a triangle will be equal. Therefore,

…… (1)

It is clear from the figure that

Now from (1)

Now as BD is tangent therefore

Therefore

From the figure we can see that

Thus we have proved.

#### Page No 8.41:

#### Question 49:

Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines .

#### Answer:

*Let ${l}_{1}\mathrm{and}{l}_{2}$ be two intersecting lines. Suppose a circle with centre O touches the lines **${l}_{1}\mathrm{and}{l}_{2}$** at M and N respectively.*

∴ OM = ON (Radius of the same circle)

⇒ O is equidistant form *${l}_{1}\mathrm{and}{l}_{2}$*.

In ΔOPM and ΔOPN,

∠OMP = ∠ONP (Radius is perpendicular to the tangent at the point of contact)

OP = OP (Common)

OM = ON (Radius of the same circle)

∴ ΔOPM ΔOPN (RHS congruence criterion)

⇒ ∠MPO = ∠NPO (CPCT)

⇒ *l* bisects ∠MPN.

⇒ O lies on the bisector of the angles between *${l}_{1}\mathrm{and}{l}_{2}$* i.e., O lies on *l*.

Thus, the centre of the circle touching two intersecting lines lies on the angle bisector of the two lines.

#### Page No 8.41:

#### Question 50:

In Fig. 8.78, there are two concentric circles with centre O. PRT and PQS are tangents to the inner circle from a point P lying on the outer circle. If PR = 5 cm, find the length of PS.

#### Answer:

Given that PR = 5 cm.

PR and PQ are the tangents to the inner circle so,

PR = PQ = 5 cm (Tangents drawn from an external point to the circle are equal)

Now draw a perpendicular from the centre O to the tangent PS.

PS is the chord of the inner circle. we know that the perpendicular drawn

from the centre of the circle to the chord bisects the chord. So, PQ = QS = 5 cm

PS = PQ + QS = 5 cm + 5 cm = 10 cm

#### Page No 8.41:

#### Question 51:

In Fig. 8.79, PQ is a tangent from an external point P to a circle with centre O and OP cuts the circle at T and QOR is a diameter. If ∠POR = 130° and S is a point on the circle, find ∠1 + ∠2.

#### Answer:

Given: ∠POR = 130°

So, ∠TSR = $\frac{1}{2}\angle \mathrm{POR}=\frac{1}{2}\times 130\xb0=65\xb0=\angle 2$ .....(1) (Since angle subtended by the arc at the centre is double the angle subtended by it at the remaining part of the circle)

∠POQ = 180º − ∠POR = 180º − 130º = 50º .....(2) (Linear pair)

In $\u25b3$POQ,

$\angle 1+\angle \mathrm{POQ}+\angle \mathrm{OQP}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle 1+50\xb0+90\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle 1=40\xb0$

$\mathrm{Now}\angle 1+\angle 2=40\xb0+65\xb0=105\xb0$

#### Page No 8.42:

#### Question 52:

In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. If PA = 12 cm, QC = QD = 3 cm, then find PC + PD.

#### Answer:

Given: PA and PB are the tangents to the circle.

PA = 12 cm

QC = QD = 3 cm

To find: PC + PD

PA = PB = 12 cm (The lengths of tangents drawn from an external point to a circle are equal)

Similarly, QC = AC = 3 cm

and QD = BD = 3 cm.

Now, PC = PA − AC = 12 − 3 = 9 cm

Similarly, PD = PB − BD = 12 − 3 = 9 cm

Hence, PC + PD = 9 + 9 = 18 cm.

#### Page No 8.45:

#### Question 1:

*Mark the correct alternative in each of the following:*

A tangent *PQ* at a point *P* of a circle of radius 5 cm meets a line through the centre *O* at a point *Q* such that *OQ* = 12 cm. Length *PQ* is

(a) 12 cm

(b) 13 cm

(c) 8.5 cm

(d) $\sqrt{119}cm$

#### Answer:

Let us first put the given data in the form of a diagram.

Given data is as follows:

*OQ* = 12 cm

*OP* = 5 cm

We have to find the length of *QP*.

We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore, *OP* is perpendicular to *QP*. We can now use Pythagoras theorem to find the length of *QP*.

Therefore the correct answer is choice (*d*).

#### Page No 8.45:

#### Question 2:

From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is

(a) 7 cm

(b) 12 cm

(c) 15 cm

(d) 24.5 cm

#### Answer:

Let us first put the given data in the form of a diagram.

The given data is as follows:

*QP* = 24 cm

*QO* = 25 cm

We have to find the length of *OP*, which is the radius of the circle.

We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore, *OP* is perpendicular to *QP*. We can now use Pythagoras theorem to find the length of *QP*.

Therefore the length of the radius of the circle is 7 cm.

Hence the correct answer to the question is choice (*a*).

#### Page No 8.45:

#### Question 3:

The length of the tangent from a point A at a circle, of radius 3 cm, is 4 cm. The distance of A from the centre of the circle is

(a) $\sqrt{7}cm$

(b) 7 cm

(c) 5 cm

(d) 25 cm

#### Answer:

Let us first put the given data in the form of a diagram.

Given data is as follows:

*OP* = 3 cm

*AP* = 4 cm

We have to find the length of *OA*.

We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore, *OP* is perpendicular to *AP*. We can now use Pythagoras theorem to find the length of *OA*.

Therefore, the distance of the point A from the center of the circle is 5 cm.

Hence the correct answer is choice (*c*).

#### Page No 8.45:

#### Question 4:

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at

(a) 50°

(b) 60°

(c) 70°

(d) 80°

#### Answer:

Let us first put the given data in the form of a diagram.

Given data is as follows:

We have to find.

We know that the radius of the circle will always be perpendicular to the tangent at the point of contact. Therefore,

Also, we know that sum of all angles of a quadrilateral will always be equal to 360°. Therefore in quadrilateral *AOBP*, we have,

Now consider *Δ**POA** *and *Δ**POB*. We have,

*PO* is the common side for both the triangles.

*PA = PB* (Length of tangents drawn from the same external point will be equal)

*OA = OB*(Radii of the same circle)

Therefore, from SSS postulate of congruent triangles,

Therefore,

We have found that

That is,

Therefore, the correct answer is choice (*a*).

#### Page No 8.45:

#### Question 5:

If TP and TQ are two tangents to a circle with centre O so that ∠POQ = 110°, then, ∠PTQ is equal to

(a) 60°

(b) 70°

(c) 80°

(d) 90°

#### Answer:

Let us first put the given data in the form of a diagram.

It is given that,

We have to find

We know that the radius of the circle will always be perpendicular to the tangent at the point of contact. Therefore,

Also, we know that sum of all angles of a quadrilateral will always be equal to 360°. Therefore in quadrilateral *TQOP*, we have,

Therefore, the correct answer to this question is choice (*b*).

#### Page No 8.45:

#### Question 6:

*PQ *is a tangent to a circle with centre *O* at the point *P*. If Δ *OPQ* is an isosceles triangle, then ∠*OQP* is equal to

(a) 30°

(b) 45°

(c) 60°

(d) 90°

#### Answer:

Let us first put the given data in the form of a diagram.

We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore, *OP* is perpendicular to *QP*. Therefore,

The side opposite tois *OQ*.*OQ* will be the longest side of the triangle. So, in the isosceles right triangle Δ*OPQ*,

*OP = PQ*

And the angles opposite to these two sides will also be equal. Therefore,

We know that sum of all angles of a triangle will always be equal to 180°. Therefore,

Therefore, the correct answer to this question is choice (*b*).

#### Page No 8.45:

#### Question 7:

Two equal circles touch each other externally at C and AB is a common tangent to the circles . Then, ∠*ACB* =

(a) 60°

(b) 45°

(c) 30°

(d) 90°

#### Answer:

Let us first put the given data in the form of a diagram.

Let us draw radius *OA*, such that it touches the circle with center *O* at point A.

Let us draw radius *O’B* such that it touches the circle with center *O’* at point *B*.

Let us also draw radii from each circle to the point *C*.

We know that the radius is always perpendicular to the tangent at the point of contact. Therefore,

That is,

…… (1)

Similarly,

Now, in ,

OA = OC(Radii of the same circle)

Therefore,

(Angles opposite to equal sides will be equal) …… (3)

Similarly, in

…… (4)

Consider the straight line *OCO’*. We have,

Looking at the figure, we can rewrite the above equation as,

(From (3) and (4))

(From (1) and (2))

…… (5)

Now let us take up. We have,

(Sum of all angles of a triangle will be 180°)

From equation (5), we get

Therefore the correct answer to this question is choice (*d*).

#### Page No 8.46:

#### Question 8:

ABC is a right angled triangle, right angled at B such that BC = 6 cm and AB = 8 cm. A circle with centre O is inscribed in Δ ABC. The radius of the circle is

(a) 1 cm

(b) 2 cm

(c) 3 cm

(d) 4 cm

#### Answer:

Let us first put the given data in the form of a diagram.

Let us first find out AC using Pythagoras theorem.

Also, we know that tangents drawn from an external point will be equal in length. Therefore we have the following,

*BL = BM* …… (1)

*CM = CN* …… (2)

*AL = AN* …… (3)

We have found that,

*AC* = 10

That is,

*AN + NC* = 10

But from (2) and (3), we can say

*AL + MC* = 10 …… (4)

It is given that,

*AB* = 8

*BC* = 6

Therefore,

*AB + BC* =14

Looking at the figure, we can rewrite this as,

*AL + LB + BM + MC* = 14

(*AL + MC*) + (*BM + BL*) = 14

Using (1) and (3) we can write the above equation as,

10 + 2*BL* = 14

*BL* = 2Consider the quadrilateral, *BLOM*, we have,

*BL = BM *(From (1))

*OL = OM*(Radii of the same circle)

(Given data)

(Radii is always perpendicular to the tangent at the point of contact)

(Radii is always perpendicular to the tangent at the point of contact)

Since the sum of all angles of a quadrilateral will be equal to , we have,

Since all the angles of the quadrilateral are equal to and since adjacent sides are equal, the quadrilateral *BLOM* is a square. We know that all the sides of a square are of equal length.

We have found *BL* = 2

Therefore,

*OM* = 2

*OM* is nothing but the radius of the circle.

Therefore, radius of the circle is 2 cm.

Hence the correct answer to the question is option (b).

#### Page No 8.46:

#### Question 9:

PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that ∠POR = 120°, then ∠OPQ is

(a) 60°

(b) 45°

(c) 30°

(d) 90°

#### Answer:

Let us first put the given data in the form of a diagram.

Given data is as follows:

*QOR* is the diameter.

=

We have to find .

Since QOR is the diameter of the circle, it is a straight line. Therefore,

That is,

But = . Therefore,

Now consider . We have

(Sum of all angles of a triangle will be )

But,

(Since radius will be perpendicular to the tangent at the point of contact)

Therefore,

Therefore, the correct answer to this question is option (c).

#### Page No 8.46:

#### Question 10:

If four sides of a quadrilateral *ABCD* are tangential to a circle, then

(a) AC + AD = BD + CD

(b) AB + CD = BC + AD

(c) AB + CD = AC + BC

(d) AC + AD = BC + DB

#### Answer:

The figure of the Quadrilateral is drawn below.

We know that length of the tangents drawn from an external point will be equal. Therefore,

*AP = PQ*

*DP = DS*

*CR = CS*

*BR = BQ*

Let us add all the above four equations. We get

*AP + DP + CR + BR = PQ + DS + CS + BQ*

By looking at the figure, we can rewrite the above equation as,

*AD + BC = AB + CD*

Therefore, the correct answer is option (c).

#### Page No 8.46:

#### Question 11:

The length of the tangent drawn from a point 8 cm away form the centre of a circle of radius 6 cm is

(a) $\sqrt{7}cm$

(b) $2\sqrt{7}$

(c) 10 cm

(d) 5 cm

#### Answer:

Let us first put the given data in the form of a diagram.

We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore, *OP* is perpendicular to *QP*. We can now use Pythagoras theorem to find the length of *QP*.

Therefore, the correct answer is option (b).

#### Page No 8.46:

#### Question 12:

AB and CD are two common tangents to circles which touch each other at C. If D lies on AB such that CD = 4 cm, then AB is equal to

(a) 4 cm

(b) 6 cm

(c) 8 cm

(d) 12 cm

#### Answer:

Let us first put the given data in the form of a diagram.

We know that tangents drawn from an external point will be equal.

Therefore,

*AD = CD*

Since CD is given as 4 cm,

*AD* = 4 cm

Similarly,

*BD = CD*

Therefore,

*BD* = 4 cm

*AB = AD + BD*

*AB* = 4 + 4

*AB* = 8 cm

Therefore, the answer to this question is option (c).

#### Page No 8.46:

#### Question 13:

In the given figure, if AD, AE and BC are tangents to the circle at D, E and F respectively, Then,

(a) AD = AB + BC + CA

(b) 2AD = AB + BC + CA

(c) 3AD = AB + BC + CA

(d) 4AD = AB + BC + CA

#### Answer:

In the given problem, the Right Hand Side of all the options is same, that is,

*AB + BC + CA*

So, we shall find out *AB + BC + CA* and check which of the options has the Left Hand Side value which we will arrive at.

By looking at the figure, we can write,

*AB + BC + CA = AB + BF + FC + CD*

We know that tangents drawn from an external point will be equal in length. Therefore,

*BF = BE*

*FC= CD*

Now we have,

*AB + BC + CA = AB + BE + CD + CA*

*AB + BC + CD = (AB + BE) + (CD + CA)*

By looking at the figure, we write the above equation as,

*AB + BC + CD = AE + AD*

Since tangents drawn from an external point will be equal,

*AE = AD*

Therefore,

*AB + BC + CD = AD + AD*

*AB + BC + CD = 2AD*

Therefore option (b) is the correct answer.

#### Page No 8.46:

#### Question 14:

In the given figure, RQ is a tangent to the circle with centre O. If SQ = 6 cm and QR = 4 cm, then OR =

(a) 8 cm

(b) 3 cm

(c) 2.5 cm

(d) 5 cm

#### Answer:

It is given that,

*SQ* = 6 cm

Since *SQ* passes through the centre of the circle *O*, it is the diameter. Therefore, the radius,

*OQ* = 3 cm

Also, given is

*QR* = 4

We know that the radius will always be perpendicular to the tangent at the point of contact. Therefore, *OQ* is perpendicular to *OR*. We can find the length of *OR* by using Pythagoras theorem. We have,

Therefore option (d) is the correct answer.

#### Page No 8.46:

#### Question 15:

In the given figure, the perimeter of Δ ABC is

(a) 30 cm

(b) 60 cm

(c) 45 cm

(d) 15 cm

#### Answer:

We know that tangents from an external point will be equal in length.

Therefore,

*AQ = AR*

*AQ* is given as 4 cm. Therefore,

*AR* = 4

Similarly,

*PC = CQ*

*PC* = 5

Therefore,

*CQ* = 5

Similarly,

*BP = BR*

*BR* = 6

Therefore,

*BP* = 6

Now we can find out the perimeter of the triangle.

Perimeter = *AB + BC + CA*

From the figure we have,

Perimeter = AR *+ RB + BP + PC + CQ + QA*

Perimeter = 4 + 6 + 6 + 5 + 5 + 4

Perimeter = 30 cm

Therefore, option (a) is the correct answer.

#### Page No 8.47:

#### Question 16:

In the given figure, AP is a tangent to the circle with centre O such that OP = 4 cm and ∠OPA = 30°. Then,

AP =

(a) $2\sqrt{2}cm$

(b) 2 cm

(c) $2\sqrt{3}$ cm

(d) $3\sqrt{2}cm$

#### Answer:

In the given figure,

if we join *O* and *A*, the line *OA* will be perpendicular to *AP*. This is because the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore, is a right triangle.

We know that,

…… (1)

We know that,

…… (2)

From equations (1) and (2), we have,

Therefore, option (c) is the correct answer.

#### Page No 8.47:

#### Question 17:

AP and PQ are tangents drawn from a point A to a circle with centre O and radius 9 cm. If OA = 15 cm, then AP + AQ =

(a) 12 cm

(b) 18 cm

(c) 24 cm

(d) 36 cm

#### Answer:

Let us first put the given data in the form of a diagram.

Since tangents drawn from an external point will be equal in length,

*AP = AQ*

Since, *AP* = 12

*AQ* = 12

*AP + AQ* = 12 + 12

*AP + AQ* = 24

Therefore option (c) is the correct answer to this question.

#### Page No 8.47:

#### Question 18:

At one end of a diameter* PQ* of a circle of radius 5 cm, tangent *XPY* is drawn to the circle. The length of chord *AB* parallel to *XY *and at a distance of 8 cm from *P* is

(a) 5 cm

(b) 6 cm

(c) 7 cm

(d) 8 cm

#### Answer:

Consider the figure.

fIG

In right angled $\u2206\mathrm{OMB}$,

By pythagoras theorem.

${\mathrm{OB}}^{2}={\mathrm{MB}}^{2}+{\mathrm{OM}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {5}^{2}={\mathrm{MB}}^{2}+{3}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 25-9={\mathrm{MB}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 16={\mathrm{MB}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 4=\mathrm{MB}$

Therefore, AM = 2MB = 2 × 4 = 8 cm.

Hence, the correct answer is option (d).

#### Page No 8.47:

#### Question 19:

If PT is tangent drawn from a point P to a circle touching it at T and O is the centre of the circle, then ∠OPT + ∠POT =

(a) 30°

(b) 60°

(c) 90°

(d) 180°

#### Answer:

Let us first put the given data in the form of a diagram.

We know that the radius will always be perpendicular to the tangent at the point of contact. Therefore,

Consider . We know that sum of all angles of a triangle will be . Therefore,

Since , we have,

Choice (c) is the right answer.

#### Page No 8.47:

#### Question 20:

In the adjacent figure, If AB = 12 cm, BC = 8 cm and AC = 10 cm, then AD =

(a) 5 cm

(b) 4 cm

(c) 6 cm

(d) 7 cm

#### Answer:

By looking at the given figure

We can write the following equations:

*AD + DB* = 12

We know that tangents form an external point will be of equal length. Therefore,

*DB = BE* …… (1)

Hence we have,

*AD + BE* = 12

From the figure, we have

*BE + EC* = 8 …… (2)

Let us subtract equation (2) from equation (3). We get,

*AD − EC* = 4 …… (3)

Since tangents from an external point will be equal, we have,

*AD = AF*

*EC = CF*

Therefore, equation (3) becomes,

*AF − CF* = 4 …… (4)

From the figure we have,

*AF + CF* = 10 …… (5)

Adding equation (4) and (5), we get,

2*AF* = 14

*AF* = 7

We know that,

*AF = AD*

Therefore,

*AD* = 7

The correct answer is option (d).

#### Page No 8.47:

#### Question 21:

In the given figure, if AP = *PB*, then

(a)

*AC*=

*AB*

(b)

*AC*=

*BC*

(c)

*AQ*=

*QC*

(d) AB = BC

#### Answer:

The given figure is below

It is given that,* *

*AP = PB*

We know that

*PB = BR*(tangents from an external point are equal)

From the above two equations, we can say that,

*AP = BR* …… (1)

Also,

*AP = AQ*(tangents from an external point are equal) …… (2)

From equations (1) and (2), we have,

*AQ = BR* …… (3)

Also,

*QC = CR*(tangents from an external point are equal) …… (4)

Adding equations (3) and (4), we get,

*AQ + QC = BR + CR*

By looking at the figure we can write the above equation as,

*AC = BC*

The correct answer is option (b).

#### Page No 8.47:

#### Question 22:

In the given figure, if AP = 10 cm, then BP =

(a) $\sqrt{91}cm$

(b) $\sqrt{127}cm$

(c) $\sqrt{119}cm$

(d) $\sqrt{109}cm$

#### Answer:

Since the radius is always perpendicular to the tangent at the point of contact,

Therefore,

Now, consider. Here also, *OB* is perpendicular to *PB* since the radius will be perpendicular to the tangent at the point of contact. Therefore,

The correct answer is option (b).

#### Page No 8.48:

#### Question 23:

In the given figure, if *PR* is tangent to the circle at *P *and *Q* is the centre of the circle, then ∠*POQ** *=

(a) 110°

(b) 100°

(c) 120°

(d) 90°

#### Answer:

We know that the radius is always perpendicular to the tangent at the point of contact.

Therefore, we have,

It is given that,

That is,

Now, consider. We have,

*OP* = *OQ* (Radii of the same circle)

Since angles opposite to equal side will be equal in a triangle, we have,

We know that sum of all angles of a triangle will be equal to .

Therefore,

The correct answer is option (c).

#### Page No 8.48:

#### Question 24:

In the given figure, if quadrilateral *PQRS* circumscribes a circle, then *PD* + *QB* =

(a)

*PQ*

(b)

*QR*

(c)

*PR*

(d)

*PS*

#### Answer:

We know that tangents drawn to a circle from the same external point will be equal in length.

Therefore,

*PD = PA* …… (1)

*QB = QA* …… (2)

Adding equations (1) and (2), we get,

*PD + QB = PA + QA*

By looking at the figure we can say,

*PD + QB = PQ*

Therefore option (a) is the correct answer.

#### Page No 8.48:

#### Question 25:

In the given figure, two equal circles touch each other at *T*, if *OP* = 4.5 cm, then *QR* =

(a) 9 cm

(b) 18 cm

(c) 15 cm

(d) 13.5 cm

#### Answer:

We know that tangents drawn from the same external point will be equal in length.

Therefore, we have,

*QP = PT*

*PT = PR*

From the above two equations, we get,

*QP = PR*

It is given that,

*QP* = 4.5 cm

Therefore,

*PR* = 4.5 cm

From the figure, we have,

*QR = QP + PR*

*QR* = 4.5 + 4.5

*QR* = 9

The correct answer is option (a).

#### Page No 8.48:

#### Question 26:

In the given figure, *APB* is a tangent to a circle with centre O at point P. If ∠*QPB* = 50°, then the measure of ∠*POQ* is

(a) 100°

(b) 120°

(c) 140°

(d) 150°

#### Answer:

We know that the radius of a circle will always be perpendicular to the tangent at the point of contact.

Therefore,

That is,

It is given that,

Therefore, we have,

Now, consider . We have,

*OP = OQ*(Radii of the same circle)

Since angles opposite to equal sides will be equal, we have,

We have found that,

Therefore,

We know that sum of all angles of a triangle will be equal to . Therefore,

The correct answer is option (a).

#### Page No 8.48:

#### Question 27:

In the given figure, if tangents *PA* and *PB* are drawn to a circle such that ∠*APB* = 30° and chord *AC* is drawn parallel to the tangent *PB*, then ∠*ABC* =

(a) 60°

(b) 90°

(c) 30°

(d) None of these

#### Answer:

We know that tangents from an external point will be equal in length. Therefore,

PA = PB

Now consider . We have,

PA = PB

We know that angles opposite to equal sides will be equal. Therefore,

Also, sum of all angles of a triangle will be equal to .

Now, as AC || PB;

Also,

The correct answer is option (c).

#### Page No 8.49:

#### Question 28:

In the given figure, *PR* =

(a) 20 cm

(b) 26 cm

(c) 24 cm

(d) 28 cm

#### Answer:

We know that the radius will always be perpendicular to the tangent at the point of contact.

Therefore,

Hence we have,

Also,

*OO’* = sum of the radii of the two circles

*OO’ *= 3 + 5

*OO’* = 8

Since radius is perpendicular to the tangent, . Therefore,

*PR = PO + OO’ + O’R*

*PR* = 5 + 8 + 13

*PR* =26

Therefore, option (b) is correct.

#### Page No 8.49:

#### Question 29:

Two circles of same radii r and centres *O* and *O*' touch each other at *P* as shown in Fig. 10.91. If *O* *O*' is produced to meet the cirele *C* (O', r) at *A* and *AT* is a tangent to the circle *C* (*O*,r) such that *O*'*Q* $\perp $ *AT*. Then *AO* : *AO*' =

(a) 3/2

(b) 2

(c) 3

(d) 1/4

#### Answer:

From the given figure we have,

*AO = r + r + r*

*AO* = 3*r*

*AO’ = r*

Therefore,

Also as therefore

Therefore, option (c) is correct.

#### Page No 8.49:

#### Question 30:

Two concentric circles of radii 3 cm and 5 cm are given. Then length of chord *BC *which touches the inner circle at *P* is equal to

(a) 4 cm

(b) 6 cm

(c) 8 cm

(d) 10 cm

#### Answer:

Consider .

We have,

Therefore,

Considering AB as the chord to the bigger circle, as OQ is perpendicular to AB, OQ bisects AB.

∴ AQ = QB = 4 cm.

Now, as BQ and BP are pair of tangents to the inner circle drawn from the external point B, QB = PB = 4 cm.

Now, join OP.

Then, _{.}

⇒ OP bisects BC

⇒ BP = PC = 4 cm

Thus, BC = 4 cm + 4 cm = 8 cm

Therefore, option (c) is correct.

#### Page No 8.49:

#### Question 31:

In the given figure, there are two concentric circles with centre O. *PR *and *PQS *are tangents to the inner circle from point plying on the outer circle. If *PR* = 7.5 cm, then *PS* is equal to

(a) 10 cm

(b) 12 cm

(c) 15 cm

(d) 18 cm

#### Answer:

Let us first draw the radii *OS*, *OP* and *OQ* in the given figure, for our convenience.

Consider . We have,

*PO = OS* (Radii of the same circle)

Therefore, is an isosceles triangle.

We know that in an isosceles triangle, if a line is drawn perpendicular to the base of the triangle from the common vertex of the equal sides, then that line will bisect the base (unequal side).

Therefore, we have

*PQ = QS*

It is given that,

*PQ* = 7.5

Therefore,

*QS* = 7.5

From the figure, we have,

*PS = PQ + QS*

*PS* = 7.5 + 7.5

*PS* = 15

Therefore, option (c) is the correct answer.

#### Page No 8.49:

#### Question 32:

In the given figure, if *AB* = 8 cm and *PE* = 3 cm, then *AE* =

(a) 11 cm

(b) 7 cm

(c) 5 cm

(d) 3 cm

#### Answer:

We know that tangents drawn from the same external point will be equal in length.

Therefore,

*AB = AC*

It is given that,

*AB* = 8 cm

Hence,

*AC* = 8 cm …… (1)

Similarly,

*PE = CE*

It is given that,

*PE* = 3

Therefore,

*CE* = 3 …… (2)

Subtracting equations (1) and (2), we get,

*AC − CE* = 8 − 3

From the figure we can see that,

*AC − CE = AE*

Therefore,

*AE* = 8 − 3

*AE* = 5 cm

Option (c) is the correct answer.

#### Page No 8.5:

#### Question 1:

Fill in the blanks:

(i) The common point of a tangent and the circle is called ..........

(ii) A circle may have .............. parallel tangents.

(iii) A tangent to a circle intersects in it ............ point(s).

(iv) A line intersecting a circle in two points is called a ...........

(v) The angle between tangent at a point on a circle and the radius through the point is ........

#### Answer:

(i) We know that the tangent to a circle is that line which touches the circle at exactly one point. This point at which the tangent touches the circle is known as the ‘point of contact’.

Therefore we have,

The common point of the tangent and the circle is called __Point of contact__.

(ii) We know that the tangent is perpendicular to the radius of the circle at the point of contact. This also means that the tangent is perpendicular to the diameter of the circle at the point of contact. The diameter of the circle can have at the most one more perpendicular line at the other end where it touches the circle. The perpendicular line at the other end of the diameter is the tangent.

Also we know that two lines which are perpendicular to a common line will be parallel to each other.

Therefore,

A circle may have __two__ parallel tangents.

(iii) From the very basic definition of tangent we know that tangent is a line that intersects the circle at exactly one point. Therefore we have,

A tangent to a circle intersects it in __one__ point.

(iv) From the definition of a secant we know that any line that intersects the circle at 2 points is a secant. Therefore, we have

A line intersecting a circle in two points is called a __secant.__

(v) One of the properties of the tangent is that it is perpendicular to the radius at the point of contact. Therefore,

The angle between the tangent at the point of contact on a circle and the radius through the point is 90°.

#### Page No 8.5:

#### Question 2:

How many tangents can a circle have?

#### Answer:

We know that circle is made of infinite points which are located at a fixed distance from a particular point. Since at each of these infinite points a tangent can be drawn, we can have infinite number of tangent for a given circle.

#### Page No 8.5:

#### Question 3:

*O* is the centre of a circle of a radius 8 cm. the tangent at a point *A* on the circle cuts a line through *O* at *B* such that *AB* = 15 cm. Find *OB*.

#### Answer:

First let us draw whatever is given in the question. This will help us understand the problem better.

Since the tangent will always be perpendicular to the radius we have drawn *OA* perpendicular to* AB*. To find the length of *OB* we have to use Pythagoras theorem.

Therefore, length of *OB* is 17 cm.

#### Page No 8.5:

#### Question 4:

If the tangent at a point *P* to a circle with centre *O *cuts a line through *O* at *Q* such that *PQ* = 24 cm and *OQ *= 25 cm. Find the radius of the circle.

#### Answer:

Let us first draw whatever is given so that we can understand the problem better.

Since the tangent to a circle is always perpendicular to the radius of the circle at the point of contact, we have drawn *OP* perpendicular to *PQ*. Thus we have a right triangle with one of its sides as the radius. To find the radius we have to use Pythagoras theorem.

Therefore the radius of the circle is 7 cm.

#### Page No 8.50:

#### Question 33:

In the given figure, *PQ* and *PR* are tangents drawn from *P* to a circle with centre *O*. If ∠*OPQ* = 35°, then

(a)

*a*= 30°,

*b*= 60°

(b)

*a*= 35°,

*b*= 55°

(c)

*a*= 40°,

*b*= 50°

(d)

*a*= 45°,

*b*= 45°

#### Answer:

Consider and . We have,

*PO* is the common side for both the triangles.

*OQ = OR*(Radii of the same circle)

*PQ = PR*(Tangents from an external point will be equal)

Therefore, from SSS postulate of congruent triangles, we have,

Therefore,

That is,

It is given that,

Therefore,

Now consider . We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. Therefore,

We know that sum of all angles of a triangle will always be equal to . Therefore,

Therefore, the correct answer is option (b).

#### Page No 8.50:

#### Question 34:

In the given figure, if *TP *and *TQ* are tangents drawn from an external point *T* to a circle with centre *O* such that ∠*TQP* = 60°, then ∠*OPQ* =

(a) 25°

(b) 30°

(c) 40°

(d) 60°

#### Answer:

Consider .

We have,

*TP = TQ*(Tangents from an external point will be equal)

We know that angles opposite to equal sides will be equal. Therefore,

It is given that,

Therefore,

We know that the radius will always be perpendicular to the tangent at the point of contact. Therefore,

Hence,

That is,

We have found that,

Therefore,

Therefore, the correct answer is option (b).

#### Page No 8.50:

#### Question 35:

In the given figure, the sides AB , BC and CA of triangle ABC, touch a circle at P, Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, then length of BC is

(a) 11 cm (b) 10 cm (c) 14 cm (d) 15 cm [CBSE 2012]

#### Answer:

It is given that the sides AB , BC and CA of ∆ABC touch a circle at P, Q and R, respectively.

Also, PA = 4 cm, PB = 3 cm and AC = 11 cm

We know that, the lengths of tangents drawn from an external point to a circle are equal.

∴ AR = AP = 4 cm

BQ = BP = 3 cm

Now, CR = AC − AR = 11 cm − 4 cm = 7 cm

∴ CQ = CR = 7 cm (Lengths of tangents drawn from an external point to a circle are equal)

Now,

BC = BQ + CQ = 3 cm + 7 cm = 10 cm

Thus, the length of BC is 10 cm.

Hence, the correct answer is option B.

#### Page No 8.50:

#### Question 36:

In the given figure, a circle touches the side DF of ΔEDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm, then the perimeter of ΔEDF is

(a) 18 cm (b) 13.5 cm (c) 12 cm (d) 9 cm [CBSE 2012]

#### Answer:

In the given figure, DH and DK are tangents drawn to the circle from an external point D.

∴ DH = DK (Lengths of tangents drawn from an external point to a circle are equal)

FH and FM are tangents drawn to the circle from an external point F.

∴ FH = FM (Lengths of tangents drawn from an external point to a circle are equal)

EK and EM are tangents drawn to the circle from an external point E.

∴ EM = EK = 9 cm (Lengths of tangents drawn from an external point to a circle are equal)

Now,

Perimeter of ΔEDF = ED + DF + EF

= ED + (DH + FH) + EF

= ED + (DK + FM) + EF (DH = DK and FH = FM)

= (ED + DK) + (FM + EF)

= EK + EM

= 9 cm + 9 cm

= 18 cm

Thus, the perimeter of ΔEDF is 18 cm.

Hence, the corrrect answer is option A.

#### Page No 8.50:

#### Question 37:

In the given figure, DE and DF are tangents from an external point D to a circle with centre A*.* If DE = 5 cm and DE ⊥ DF, then the radius of the circle is

(a) 3 cm (b) 5 cm (c) 4 cm (d) 6 cm [CBSE 2013]

#### Answer:

It is given that DE and DF are tangents from an external point D to a circle with centre A*. *DE = 5 cm and DE ⊥ DF.

Join AE and AF.

Now, DE is a tangent at E and AE is the radius through the point of contact E.

$\therefore \angle \mathrm{AED}=90\xb0$ (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

Also, DF is a tangent at F and AF is the radius through the point of contact F.

$\therefore \angle \mathrm{AFD}=90\xb0$ (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

$\angle \mathrm{EDF}=90\xb0$ (DE ⊥ DF)

Also, DF = DE (Lengths of tangents drawn from an external point to a circle are equal)

So, AEDF is a square.

∴ AE = AF = DE = 5 cm (Sides of square are equal)

Thus, the radius of the circle is 5 cm.

Hence, the correct answer is option B.

#### Page No 8.51:

#### Question 38:

In the given figure, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches sides BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90° and DS = 5 cm, then the radius of the circle (in cm) is

(a) 11 (b) 18 (c) 6 (d) 15 [CBSE 2013]

#### Answer:

It is given that the sides BC, AB, AD and CD touches a circle with centre O at points P, Q, R and S, respectively.

AB = 29 cm, AD = 23 cm, ∠B = 90° and DS = 5 cm.

Now,

DR = DS = 5 cm (Lengths of tangents drawn from an external point to a circle are equal)

AR = AD − DR = 23 cm − 5 cm = 18 cm

∴ AQ = AR = 18 cm (Lengths of tangents drawn from an external point to a circle are equal)

BQ = AB − AQ = 29 cm − 18 cm = 11 cm

∴ BP = BQ = 11 cm (Lengths of tangents drawn from an external point to a circle are equal)

Now, AB is a tangent at Q and OQ is the radius through the point of contact Q.

$\therefore \angle \mathrm{OQB}=90\xb0$ (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

BC is a tangent at P and OP is the radius through the point of contact P.

$\therefore \angle \mathrm{OPB}=90\xb0$ (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

$\angle \mathrm{B}=90\xb0$ (Given)

Also, BP = BQ

So, OPBQ is a square.

∴ OP = OQ = BQ = 11 cm (Sides of square are equal)

Thus, the radius of the circle is 11 cm.

Hence, the correct answer is option A.

#### Page No 8.51:

#### Question 39:

In a right triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle (in cm) is

(a) 4 (b) 3 (c) 2 (d) 1 [CBSE 2014]

#### Answer:

Let a circle with centre O and radius *r* cm be inscribed in the right ∆ABC.

In right ∆ABC,

${\mathrm{AC}}^{2}={\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}\left(\mathrm{Pythagoras}\mathrm{theorem}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=\sqrt{{\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=\sqrt{{5}^{2}+{12}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=\sqrt{169}=13\mathrm{cm}$

Now, the sides AB, BC and CA are tangents to the circle at R, P and Q, respectively.

∴ OR ⊥ AB, OP ⊥ BC and OQ ⊥ AC

Now,

ar(∆OAB) + ar(∆OBC) + ar(∆OAC) = ar(∆ABC)

$\Rightarrow \frac{1}{2}\times \mathrm{AB}\times \mathrm{OR}+\frac{1}{2}\times \mathrm{BC}\times \mathrm{OP}+\frac{1}{2}\times \mathrm{AC}\times \mathrm{OQ}=\frac{1}{2}\times \mathrm{BC}\times \mathrm{AB}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\times 5\times r+\frac{1}{2}\times 12\times r+\frac{1}{2}\times 13\times r=\frac{1}{2}\times 12\times 5\phantom{\rule{0ex}{0ex}}\Rightarrow 5r+12r+13r=60\phantom{\rule{0ex}{0ex}}\Rightarrow 30r=60\phantom{\rule{0ex}{0ex}}\Rightarrow r=2$

Thus, the radius of the circle is 2 cm.

Hence, the correct answer is option C.

#### Page No 8.51:

#### Question 40:

Two circles touch each other externally at P. AB is a common tangent to the circle touching them at A and B. The value of $\angle $APB is

(a) 30º (b) 45º (c) 60º (d) 90º [CBSE 2014]

#### Answer:

It is given that two circles touch each other externally at P. AB is a common tangent to the circles touching them at A and B.

Draw a tangent to the circles at P, intersecting AB at T.

Now, TA and TP are tangents drawn to the same circle from an external point T.

∴ TA = TP (Lengths of tangents drawn from an external point to a circle are equal)

TB and TP are tangents drawn to the same circle from an external point T.

∴ TB = TP (Lengths of tangents drawn from an external point to a circle are equal)

In ∆ATP,

TA = TP

$\therefore \angle \mathrm{APT}=\angle \mathrm{PAT}$ .....(1) (In a triangle, equal sides have equal angles opposite to them)

In ∆BTP,

TB = TP

$\therefore \angle \mathrm{BPT}=\angle \mathrm{PBT}$ .....(2) (In a triangle, equal sides have equal angles opposite to them)

Now, in ∆APB,

$\angle \mathrm{APB}+\angle \mathrm{PAB}+\angle \mathrm{PBA}=180\xb0$ (Angle sum property)

$\Rightarrow \angle \mathrm{APB}+\angle \mathrm{APT}+\angle \mathrm{BPT}=180\xb0\left[\mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{APB}+\angle \mathrm{APB}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 2\angle \mathrm{APB}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{APB}=90\xb0$

Thus, the value of $\angle $APB is 90º.

Hence, the correct answer is option D.

#### Page No 8.51:

#### Question 41:

In the given figure, PQ and PR are two tangents to a circle with centre O. If $\angle $QPR = 46º, then $\angle $QOR is

(a) 67º (b) 134º (c) 44º (d) 46º [CBSE 2014]

#### Answer:

It is given that PQ and PR are two tangents to a circle with centre O.

Now, PQ is a tangent at Q and OQ is the radius through the point of contact Q.

∴ $\angle $OQP = 90º (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

PR is a tangent at R and OR is the radius through the point of contact R.

∴ $\angle $ORP = 90º (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

Also, $\angle $QPR = 46º (Given)

In quadrilateral OQPR,

$\angle $OQP + $\angle $QPR + $\angle $ORP + $\angle $QOR = 360º (Angle sum property)

⇒ 90º + 46º + 90º + $\angle $QOR = 360º

⇒ 226º + $\angle $QOR = 360º

⇒ $\angle $QOR = 360º − 226º = 134º

∴ $\angle $QOR = 134º

Hence, the correct answer is option B.

#### Page No 8.51:

#### Question 42:

In the given figure, QR is a common tangent to the given circles touching externally at the point T. The tangent at T meets QR at P. If PT = 3.8 cm, then the length of QR (in cm) is

(a) 3.8 (b) 7.6 (c) 5.7 (d) 1.9 [CBSE 2014]

#### Answer:

It is given that QR is a common tangent to the given circles touching externally at the point T. Also, the tangent at T meets QR at P such that PT = 3.8 cm.

Now, PQ and PT are tangents drawn to the same circle from an external point P.

∴ PQ = PT = 3.8 cm (Lengths of tangents drawn from an external point to a circle are equal)

PR and PT are tangents drawn to the same circle from an external point T.

∴ PR = PT = 3.8 cm (Lengths of tangents drawn from an external point to a circle are equal)

Now,

QR = PQ + PR = 3.8 cm + 3.8 cm = 7.6 cm

Thus, the length of QR is 7.6 cm.

Hence, the correct answer is option B.

#### Page No 8.51:

#### Question 43:

In the given figure, a quadrilateral ABCD is drawn to circumscribe a circle such that its sides AB, BC, CD and AD touch the circle at P, Q, R and S respectively. If AB = *x* cm, BC = 7 cm, CR = 3 cm and AS = 5 cm, then *x* =

(a) 10 (b) 9 (c) 8 (d) 7 [CBSE 2014]

#### Answer:

It is given that a quadrilateral ABCD is drawn to circumscribe a circle such that its sides AB, BC, CD and AD touch the circle at P, Q, R and S, respectively.

Also, BC = 7 cm, CR = 3 cm and AS = 5 cm

Now, AP and AS are tangents drawn to the same circle from an external point A.

∴ AP = AS = 5 cm .....(1) (Lengths of tangents drawn from an external point to a circle are equal)

CQ and CR are tangents drawn to the same circle from an external point C.

∴ CQ = CR = 3 cm (Lengths of tangents drawn from an external point to a circle are equal)

So, BQ = BC − CQ = 7 cm − 3 cm = 4 cm

BP and BQ are tangents drawn to the same circle from an external point B.

BP = BQ = 4 cm .....(2) (Lengths of tangents drawn from an external point to a circle are equal)

Now,

AB = AP + PB = 5 cm + 4 cm = 9 cm [From (1) and (2)]

∴ *x* = 9 (AB = *x* cm)

Hence, the correct answer is option B.

#### Page No 8.52:

#### Question 44:

If angle between two radii of a circle is 130^{0 }, the angle between the tangents at the ends of radii is

(a) 90^{0} (b) 50^{0} (c) 70^{0} (d) 40^{0}

#### Answer:

Let PQ and RP be the radii of the circle with the centre O.

$\angle \mathrm{ROQ}=130\xb0$

$\mathrm{RP}\perp \mathrm{OR}\mathrm{and}\mathrm{PQ}\perp \mathrm{OQ}$ (Radii are perpendicular to the tangent)

In quadrilateral ROQP,

$\angle \mathrm{ORP}+\angle \mathrm{RPQ}+\angle \mathrm{PQO}+\angle \mathrm{QOR}=360\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 90\xb0+\angle \mathrm{RPQ}+90\xb0+130\xb0=360\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{RPQ}=50\xb0$

Hence, the correct answer is option (b).

#### Page No 8.52:

#### Question 45:

If two tangents inclined at an angle of 60^{0 } are drawn to a circle of radius 3 cm , then length of each tangent is equal to

(a) $\frac{3\sqrt{3}}{2}$ (b) 6 cm (c) 3 cm (d) $3\sqrt{3}$ cm

#### Answer:

Let PA and PB be two tangents to a circle with centre O and radius 3 cm.

We are given ∠APB = 60°

We know that two tangents drawn to a circle from an external point are equally inclined to the segment joining the centre to the point.

∴ ∠APO = ∠BPO = $\frac{1}{2}$ × ∠APB = $\frac{1}{2}$ × 60° = 30°

Also, OA ⊥ AP and OB ⊥ BP (radius ⊥ tangent at point of contact)

In right ΔOAP,

$\mathrm{tan}30\xb0=\frac{3}{PA}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{\sqrt{3}}=\frac{3}{PA}\phantom{\rule{0ex}{0ex}}\Rightarrow PA=3\sqrt{3}$

Therefore, PA = PB = $3\sqrt{3}\mathrm{cm}$

Hence, the correct answer is option (d).

#### Page No 8.52:

#### Question 46:

If radii of two concentric circles are 4 cm and 5 cm , then the length of each chord of one circle which is tangent to the other circle is

(a) 3 cm (b) 6 cm (c) 9 cm (d) 1 cm

#### Answer:

Let the chord of one circle which tangent to the other be AB.

In ∆AOC,

${\mathrm{OC}}^{2}+{\mathrm{AC}}^{2}={\mathrm{OA}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}={\mathrm{OA}}^{2}-{\mathrm{OC}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}={5}^{2}-{4}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}=25-16=9\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=3\mathrm{cm}$

thus, AB = AC + CB = 3 + 3 = 6 cm

Hence, the correct answer is option (b).

#### Page No 8.52:

#### Question 47:

At one end *A* of a diameter *AB* of a circle of radius 5 cm , tangent *XAY* is drawn to the circle. The length of the chord *CD* parallel to* XY *and at a distance 8 cm from *A* is

(a) 4 cm (b) 5 cm (c) 6 cm (d) 8 cm

#### Answer:

XY is the tangent to the circle with centre O.

CD is the chord.

OA = OB = OD = 5 cm (Radii)

PA = 8 cm

PO = 3 cm

In ∆POD,

${\mathrm{PD}}^{2}+{\mathrm{PO}}^{2}={\mathrm{OD}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {3}^{2}+{\mathrm{PD}}^{2}={5}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{PD}}^{2}=25-9=16\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{PD}=4\mathrm{cm}$

Hence, CD = CP + PD = 4 + 4 = 8 cm

so, the correct answer is option (d).

#### Page No 8.52:

#### Question 48:

From a point *P *which is at a distance 13 cm from the centre *O* of a circle of radious 5 cm , the pair of tangents *PQ* and *PR *to the circle are drawn . Then the area of the quadrilateral *PQOR* is

(a) 60 cm^{2 } (b) 65 cm^{2 } (c) 30 cm^{2 } (d) 32.5 cm^{2 }

#### Answer:

Clearly tangent PQ and PR are perpendicular to OQ and OR respectively.

Hence both triangles POQ and PQR are right angled.

$\mathrm{PQ}=\sqrt{{\mathrm{OP}}^{2}-{\mathrm{OQ}}^{2}}=\sqrt{{13}^{2}-{5}^{2}}=12\mathrm{cm}$

Area of ∆POQ = $\frac{\mathrm{OQ}\times \mathrm{PQ}}{2}=\frac{5\times 12}{2}=30{\mathrm{cm}}^{2}$

Similarly,

Area of ∆POR = Area of ∆POQ = 30 cm^{2 }(Both the triangles are symmetrical)

Area of quadrilateral PQOR = 30 + 30 = 60 cm^{2}

Hence, the correct answer is option (a).

#### Page No 8.52:

#### Question 49:

If *PA* and *PB *are tangents to the circle with centre *O *such that $\angle $*APB* = 50^{0}, then $\angle $*OAB* is equal to

(a) 25^{0} (b) 30^{0} (c) 40^{0 } (d) 50^{0}

#### Answer:

In quadrilateral APBO,

$\angle APB=50\xb0$

$\angle OAP+\angle APB+\angle PBO+\angle AOB=360\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 90\xb0+90\xb0+50\xb0+\angle AOB=360\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle AOB=130\xb0$

In ∆AOB,

Let $\angle OAB=\angle OBA=x$ (OA = OB as they are radii)

$\angle AOB+\angle OBA+\angle OAB=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 130\xb0+2x=180\phantom{\rule{0ex}{0ex}}\Rightarrow x=25\xb0$

Hence, the correct answer is option (a).

#### Page No 8.52:

#### Question 50:

The pair of tangents *AP *and* AQ *drawn from an external point to a circle with centre *O* are perpendicular to each other and length of each tangent is 5 cm. The radius of the circle is

(a) 10 cm (b) 7.5 cm (c) 5cm (d) 2.5 cm

#### Answer:

Given:* AP and AQ are tangents to the circle with centre O, AP ⊥ AQ*

and AP = AQ = 5 cm

we know that radius of a circle is perpendicular to the tangent at the point of contact

⇒ OP ⊥ AP and OQ ⊥ AQ

Also sum of all angles of a quadrilateral is 360°

⇒∠O + ∠P + ∠A + ∠Q = 360°

⇒∠O + 90° + 90° + 90° = 360°

⇒∠O = 360° – 270° = 90°

Thus ∠O = ∠P = ∠A = ∠Q = 90°

⇒ OPAQ is a rectangle

Since adjacent sides of OPAQ i.e. AP and AQ are equal. Thus OPAQ is a square

radius = OP = OQ = AP = AQ = 5 cm

Hence, the correct answer is option (c).

#### Page No 8.52:

#### Question 51:

In Fig.10.114, if $\angle AOB={125}^{\xb0}$ , then $\angle COD$ is equal to

figure

(a) 45^{0} (b) 35^{0} (c) 55^{0 } (d) $62{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{\xb0}$

#### Answer:

We know that opposite sides of a quadrilateral circumscribing a circle subtend supplementary

angles at the centre.

Quadrilateral ABCD circumscribe a circle with centre O.

$\angle AOB+\angle COD=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 125\xb0+\angle COD=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle COD=180\xb0-125\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle COD=55\xb0$

Hence, the correct answer is option (c).

#### Page No 8.53:

#### Question 52:

In Fig . 10.115, if *PQR *is the tangent to a circle at Q whose centre is O , *AB* is a chord parallel to *PR *and $\angle BQR={70}^{\xb0}$, then $\angle AQB$ is equal to

figure

(a) 20^{0 } (b) 40^{0 } 9c) 35^{0} (d) 45^{0}

0

#### Answer:

Given: PQR is the tangent to the circle with centre O.

$\angle \mathrm{BQR}=70\xb0$

$\angle \mathrm{ABQ}=\angle \mathrm{BQR}=70\xb0\left(\mathrm{Alternate}\mathrm{interior}\mathrm{angles}\right)$

$\angle \mathrm{BSQ}=90\xb0$ (radius is perpendicular to the chord)

In ∆BSQ,

$70\xb0+\angle \mathrm{BQS}+90\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{BQS}=20\xb0$

Similarly, $\angle \mathrm{AQS}=20\xb0$

Thus,

$\angle \mathrm{BQS}+\angle \mathrm{AQS}=\angle \mathrm{AQB}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{AQB}=20\xb0+20\xb0=40\xb0$

Hence, the correct answer is option (b).

#### Page No 8.53:

#### Question 1:

A line intersecting a circle in two distinct points is called a __________.

#### Answer:

A line intersecting a circle in two distinct points is called a ___secant___.

#### Page No 8.53:

#### Question 2:

The tangents drawn, at the end-points of a diameter, to a circle are ___________.

#### Answer:

AB is the diameter of the circle with centre O. CD and EF are the tangents to the circle at points A and B, respectively.

Now,

∠OAD = 90º (Radius is perpendicular to the tangent at the point of contact)

Also, ∠OBF = 90º (Radius is perpendicular to the tangent at the point of contact)

∴ ∠OAD + ∠OBF = 90º + 90º = 180º

⇒ CD || EF (If a pair of consecutive interior angles is supplementary, then the two lines are parallel)

Hence, the tangents drawn at the end-points of a diameter to a circle are parallel.

The tangents drawn, at the end-points of a diameter, to a circle are ____parallel____.

#### Page No 8.53:

#### Question 3:

The lengths of two tangents drawn from an external point to a circle are _________.

#### Answer:

PA and PB are two tangents drawn from P to circle with centre O. OP is the line segment joining the point P and centre of the circle.

In ∆OAP and ∆OBP,

OA = OB (Radius of the circle)

OP = OP (Common)

∠OAP = ∠OBP (Radius is perpendicular to the tangent at the point of contact)

∴ ∆OAP $\cong $ ∆OBP (RHS congruence criterion)

⇒ PA = PB (CPCT)

Hence, the lengths of two tangents drawn from an external point to a circle are equal.

The lengths of two tangents drawn from an external point to a circle are ____equal____.

#### Page No 8.53:

#### Question 4:

The two tangents drawn from an external point to a circle are _______ to the segment joining the centre to the point.

#### Answer:

PA and PB are two tangents drawn from P to circle with centre O. OP is the line segment joining the point P and centre of the circle.

In ∆OAP and ∆OBP,

OA = OB (Radius of the circle)

OP = OP (Common)

∠OAP = ∠OBP (Radius is perpendicular to the tangent at the point of contact)

∴ ∆OAP $\cong $ ∆OBP (RHS congruence criterion)

⇒ ∠APO = ∠BPO (CPCT)

Hence, the two tangents drawn from an external point to a circle are equally inclined to the segment joining the centre to the point.

The two tangents drawn from an external point to a circle are ___equally inclined___ to the segment joining the centre to the point.

#### Page No 8.53:

#### Question 5:

Tangents drawn from an external point to a circle subtend ________ angles to the centre.

#### Answer:

PA and PB are two tangents drawn from P to circle with centre O. OP is the line segment joining the point P and centre of the circle.

In ∆OAP and ∆OBP,

OA = OB (Radius of the circle)

OP = OP (Common)

∠OAP = ∠OBP (Radius is perpendicular to the tangent at the point of contact)

∴ ∆OAP $\cong $ ∆OBP (RHS congruence criterion)

⇒ ∠AOP = ∠BOP (CPCT)

Hence, the tangents drawn from an external point to a circle subtend equal angles to the centre.

Tangents drawn from an external point to a circle subtend ____equal____ angles to the centre.

#### Page No 8.53:

#### Question 6:

Parallelogram circumscribing a circle is a _______.

#### Answer:

ABCD is a parallelogram such that its sides touches the circle with centre O at P, Q, R and S.

We know that the opposite sides of the parallelogram are equal. Therefore,

AB = CD .....(1)

BC = AD .....(2)

Also, the length of the tangents drawn from an external point to a circle are equal.

∴ AR = AS .....(3)

DR = DQ .....(4)

BP = SB .....(5)

CP = CQ .....(6)

Adding (3), (4), (5) and (6), we get

AR + DR + BP + CP = AS + SB + CQ + DQ

⇒ AD + BC = AB + CD

⇒ 2AD = 2AB [From (1) and (2)]

⇒ AD = AB .....(7)

From (1), (2) and (7), we have

AB = BC = CD = AD

Hence, the parallelogram ABCD is a rhombus.

Parallelogram circumscribing a circle is a ___rhombus___.

#### Page No 8.53:

#### Question 7:

If *AB* and *CD* are common tangents to two circles of unequal radii, then _________.

#### Answer:

Here, AB and CD are common tangents to two circles of unequal radii and having centres O_{1} and O_{2}.

AB and CD are produced to intersect at P.

Now, PB and PD are tangents drawn from external point P to circle with centre O_{2}.

∴ PB = PD .....(1) (Lengths of tangents drawn from an external point to a circle are equal)

Also, PA and PC are tangents drawn from external point P to circle with centre O_{1}.

∴ PA = PC .....(2) (Lengths of tangents drawn from an external point to a circle are equal)

Subtracting (1) from (2), we get

PA − PB = PC − PD

⇒ AB = CD

If AB and CD are common tangents to two circles of unequal radii, then ____AB = CD____.

#### Page No 8.53:

#### Question 8:

From an external point *P*, two tangents *PA *and *PB *are drawn to the circle with centre *O*. Then the angle between *OP *and *AB *is_______.

#### Answer:

PA and PB are tangents drawn from P to circle with centre O. Let OP and AB intersect at Q.

In ∆PAQ and ∆PBQ,

AP = BP (Length of tangents drawn from an external point to a circle are equal)

∠APQ = ∠BPQ (Tangents drawn from an external point to a circle are equally inclined to the segment joining the centre to that point)

PQ = PQ (Common)

∴ ∆PAQ $\cong $ ∆PBQ (SAS congruence axiom)

⇒ ∠AQP = ∠BQP (CPCT)

Also,

∠AQP + ∠BQP = 180º (Linear pair of angles)

⇒ 2∠AQP = 180º (∠AQP = ∠BQP)

⇒ ∠AQP = 90º

Therefore, OP is perpendicular to AB i.e. the angle between OP and AB is 90º.

From an external point P, two tangents PA and PB are drawn to the circle with centre O. Then the angle between OP and AB is ____90º____.

#### Page No 8.53:

#### Question 10:

*AB *is a diameter of a circle and *AC *is its chord such that ∠*BAC *= 30°. If the tangent at *C* intersects *AB *extended at *D, *then *BC *= ________.

#### Answer:

AB is a diameter of circle with centre O and AC is the chord of the circle such that ∠BAC = 30°. The tangent at C intersects AB extended at D.

In ∆OAC,

OA = OC (Radius of circle)

∴ ∠OCA = ∠OAC (In a triangle, equal sides have equal angles opposite to them)

⇒ ∠OCA = 30°

Now, ∠OCD = 90° (Radius is perpendicular to the tangent at the point of contact)

∴ ∠ACD = ∠OCA + ∠OCD = 30° + 90° = 120°

In ∆ACD,

∠A + ∠ACD + ∠D = 180º (Angle sum property)

⇒ 30° + 120° + ∠D = 180º

⇒ ∠D = 180º − 150° = 30° .....(1)

Now, ∠ACB = 90° (Angle in a semi-circle is 90°)

∠ACD = ∠ACB + ∠BCD

⇒ 120° = 90° + ∠BCD

⇒ ∠BCD = 120º − 90° = 30° .....(2)

In ∆BCD,

∠D = ∠BCD [From (1) and (2)]

⇒ BC = BD (In a triangle, equal angles have equal sides opposite to them)

AB is a diameter of a circle and AC is its chord such that ∠BAC = 30°. If the tangent at C intersects AB extended at D, then BC = ___ BD ___.

#### Page No 8.53:

#### Question 11:

If a chord *AB *subtends an angle of 60° at the centre of a circle, then the angle between the tangents at *A* and *B* is ________.

#### Answer:

AB is the chord of a circle with centre O such that ∠AOB = 60º. The tangents at A and B intersect at P.

Now,

∠OAP = 90º (Radius is perpendicular to the tangent at the point of contact)

Also, ∠OBP = 90º (Radius is perpendicular to the tangent at the point of contact)

In quadrilateral OAPB,

∠AOB + ∠OAP + ∠OBP + ∠APB = 360º (Angle sum property of quadrilateral)

⇒ 60º + 90º + 90º + ∠APB = 360º

⇒ ∠APB = 360º − 240º = 120º

Thus, the angle between the tangents at A and B is 120º.

If a chord AB subtends an angle of 60° at the centre of a circle, then the angle between the tangents at A and B is ____120º____.

#### Page No 8.54:

#### Question 12:

The length of the tangent from an external point *P* on the circle with centre *O* is always ________ *OP*.

#### Answer:

PT is a tangent drawn from an external point P on the circle with centre O.

∠OTP = 90º (Radius is perpendicular to the tangent at the point of contact)

So, ∆OPT is a right triangle. OP is the hypotenuse of the right ∆OPT.

∴ OP > PT (In a right triangle, hypotenuse is the longest side.)

Or PT < OP

Thus, the length of the tangent from an external point P on the circle with centre O is always less than the hypotenuse OP.

The length of the tangent from an external point P on the circle with centre O is always ____less than____ OP.

#### Page No 8.54:

#### Question 13:

If the angle between two tangents drawn from a point *P* to a circle of radius *a *and centre *O* is 60°, then *OP *= _______.

#### Answer:

PA and PB are tangents drawn from P to circle with centre O.

∠APB = 60º and OA = OB = *a* (Radius of the circle)

We know that, the two tangents drawn from an external point to a circle are equally inclined to the segment joining the centre to the point.

∴ ∠APO = ∠BPO = $\frac{60\xb0}{2}$ = 30º

Also, ∠OAP = 90º (Radius is perpendicular to the tangent at the point of contact)

In right ∆OAP,

$\mathrm{sin}30\xb0=\frac{\mathrm{OA}}{\mathrm{OP}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}=\frac{a}{\mathrm{OP}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OP}=2a$

If the angle between two tangents drawn from a point P to a circle of radius *a* and centre O is 60°, then OP = _____2 a_____.

#### Page No 8.54:

#### Question 14:

If a number of circles pass through the end points *P* and *Q *of a line segment *PQ*, then their centres lie on the perpendicular bisector of _______.

#### Answer:

Consider two circles with centres O_{1} and O_{2} passing through the end points P and Q of a line segment PQ.

We know that the perpendicular bisector of the chord of a circle passes through the centre of the circle.

Thus, the perpendicular bisector PQ passes through the centres O_{1} and O_{2}.

In the same manner, all circles passing through the end points P and Q of a line segment PQ will have their centres lying on the perpendicular bisector of PQ.

If a number of circles pass through the end points P and Q of a line segment PQ, then their centres lie on the perpendicular bisector of _____PQ_____.

#### Page No 8.54:

#### Question 15:

The angle between two tangents drawn from an external point to a circle is _______ to the angle subtended by the line segments joining the points of contact at the centre.

#### Answer:

PA and PB are two tangents drawn from P to circle with centre O.

Now,

∠OAP = 90º (Radius is perpendicular to the tangent at the point of contact)

Also, ∠OBP = 90º (Radius is perpendicular to the tangent at the point of contact)

In quadrilateral OAPB,

∠AOB + ∠OAP + ∠OBP + ∠APB = 360º (Angle sum property of quadrilateral)

⇒ ∠AOB + 90º + 90º + ∠APB = 360º

⇒ ∠AOB + ∠APB = 360º − 180º = 180º

Thus, the angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact at the centre.

The angle between two tangents drawn from an external point to a circle is ____supplementary____ to the angle subtended by the line segments joining the points of contact at the centre.

#### Page No 8.54:

#### Question 16:

*PA *and *PB *are two tangents drawn from an external point* P *to a circle with centre *O*. If ∠*PBA *= 65°, then ∠*APB *= _______.

#### Answer:

PA and PB are two tangents drawn from an external point P to a circle with centre O.

∴ PA = PB (Lengths of tangents drawn from an external point to a circle are equal)

In ∆APB,

PB = PA (Proved)

⇒ ∠PAB = ∠PBA (In a triangle, equal sides have equal angles opposite to them)

⇒ ∠PAB = 65° (∠PBA = 65°)

Also,

∠PAB + ∠PBA + ∠APB = 180° (Angle sum property of triangle)

⇒ 65° + 65° + ∠APB = 180°

⇒ ∠APB = 180° − 130° = 50°

PA and PB are two tangents drawn from an external point P to a circle with centre O. If ∠PBA = 65°, then ∠APB = _____50°_____.

#### Page No 8.54:

#### Question 17:

*PA *and *PB *are two tangents drawn from an external point *P* to a circle with centre *O*. If ∠*APB *= 80°, then ∠*POA *= __________.

#### Answer:

PA and PB are tangents drawn from P to circle with centre O.

∠APB = 80º

We know that, the two tangents drawn from an external point to a circle are equally inclined to the segment joining the centre to the point.

∴ ∠APO = ∠BPO = $\frac{80\xb0}{2}$ = 40º

Also, ∠OAP = 90º (Radius is perpendicular to the tangent at the point of contact)

In ∆OAP,

∠APO + ∠OAP + ∠POA = 180º (Angle sum property of triangle)

⇒ 40º + 90º + ∠POA = 180º

⇒ ∠POA = 180º − 130º = 50º

PA and PB are two tangents drawn from an external point P to a circle with centre O. If ∠APB = 80°, then ∠POA = ______50°______.

#### Page No 8.54:

#### Question 18:

If *PA *and *PB *are two tangents drawn from an external point *P *to a circle such that *PA *= 5 cm and ∠*APB *= 60°, then the length of chord *AB *is ________.

#### Answer:

PA and PB are tangents drawn from P to circle with centre O. Let OP and AB intersect at Q.

∠APQ = ∠BPQ = $\frac{60\xb0}{2}$ = 30º (Tangents drawn from an external point to a circle are equally inclined to the segment joining the centre to that point)

In ∆PAQ and ∆PBQ,

AP = BP (Length of tangents drawn from an external point to a circle are equal)

∠APQ = ∠BPQ (30º each)

PQ = PQ (Common)

∴ ∆PAQ $\cong $ ∆PBQ (SAS congruence axiom)

⇒ ∠AQP = ∠BQP and AQ = BQ (CPCT)

Also,

∠AQP + ∠BQP = 180º (Linear pair of angles)

⇒ 2∠AQP = 180º (∠AQP = ∠BQP)

⇒ ∠AQP = 90º

In right ∆APQ,

$\mathrm{sin}30\xb0=\frac{\mathrm{AQ}}{\mathrm{AP}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}=\frac{\mathrm{AQ}}{5\mathrm{cm}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AQ}=\frac{5}{2}\mathrm{cm}$

∴ AB = 2AQ = $2\times \frac{5}{2}$ = 5 cm (AQ = BQ)

Thus, the length of the chord AB is 5 cm.

If PA and PB are two tangents drawn from an external point P to a circle such that PA = 5 cm and ∠APB = 60°, then the length of chord AB is _____5 cm_____.

#### Page No 8.54:

#### Question 19:

Quadrilateral *ABCD *is circumscribed to a circle. If *AB *= 6 cm, *BC** = *7 cm and* **CD *= 4 cm, then *AD *= ________.

#### Answer:

ABCD is a quadrilateral such that its sides touches the circle with centre O at P, Q, R and S.

We know that the lengths of the tangents drawn from an external point to a circle are equal. Therefore,

AR = AS .....(1)

DR = DQ .....(2)

BP = SB .....(3)

CP = CQ .....(4)

Adding (1), (2), (3) and (4), we get

AR + DR + BP + CP = AS + SB + CQ + DQ

⇒ AD + BC = AB + CD

⇒ AD + 7 cm = 6 cm + 4 cm

⇒ AD = 10 − 7 = 3 cm

Quadrilateral ABCD is circumscribed to a circle. If AB = 6 cm, BC = 7 cm and CD = 4 cm, then AD = ____3 cm____.

#### Page No 8.54:

#### Question 20:

The length of the tangent from an external point *P* to a circle of radius 5 cm is 10 cm. The distance of the point from the centre of the circle is _______.

#### Answer:

PT is a tangents drawn from P to circle with centre O.

PT = 10 cm and OT = 5 cm (Radius of the circle)

Now,

∠OTP = 90º (Radius is perpendicular to the tangent at the point of contact)

In right ∆OTP,

${\mathrm{OP}}^{2}={\mathrm{OT}}^{2}+{\mathrm{PT}}^{2}\left(\mathrm{Pythagoras}\mathrm{theorem}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{OP}}^{2}={\left(5\right)}^{2}+{\left(10\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{OP}}^{2}=25+100=125\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OP}=5\sqrt{5}\mathrm{cm}$

Thus, the distance of the point from the centre of the circle is $5\sqrt{5}$ cm.

The length of the tangent from an external point *P* to a circle of radius 5 cm is 10 cm. The distance of the point from the centre of the circle is $\overline{)5\sqrt{5}\mathrm{cm}}$.

#### Page No 8.54:

#### Question 1:

In the given figure, *PA* and *PB* are tangents to the circle drawn from an external point P. *CD* is a third tangent touching the circle at Q. If *PB* = 10 cm and *CQ* = 2 cm, what is the length *PC*?

#### Answer:

Given data is as follows:

PB = 10 cm

CQ = 2 cm

We have to find the length of PC.

We know that the length of two tangents drawn from the same external point will equal. Therefore,

PB = PA

It is given that PB = 10 cm

Therefore, PA = 10 cm

Also, from the same principle we have,

CQ = CA

It is given that CQ = 2 cm

Therefore, CA = 2cm

From the given figure we can say that,

PC = PA − CA

Now that we know the values of PA and CA, let us substitute the values in the above equation.

PC = 10 − 2

PC = 8 cm

Therefore, length of PC is 8 cm.

#### Page No 8.54:

#### Question 2:

What is the distance between two parallel tangents of a circle of radius 4 cm?

#### Answer:

Two parallel tangents can exist at the two ends of the diameter of the circle. Therefore, the distance between the two parallel tangents will be equal to the diameter of the circle. In the problem the radius of the circle is given as 4 cm. Therefore,

Diameter =

Diameter = 8 cm

Hence, the distance between the two parallel tangents is 8 cm.

#### Page No 8.54:

#### Question 3:

The length of tangent from a point A at a distance of 5 cm from the centre of the circle is 4 cm. What is the radius of the circle?

#### Answer:

Let us first draw whatever is given for a better understanding of the problem.

Let *O* be the center of the circle and *B* be the point of contact.

We know that the radius of the circle will be perpendicular to the tangent at the point of contact. Therefore, we have as a right triangle and we have to apply Pythagoras theorem to find the radius of the triangle.

Therefore, radius of the circle is 3 cm.

#### Page No 8.55:

#### Question 4:

Two tangents TP and TQ are drawn from an external point T to a circle with centre O as shown in Fig. 10.73. If they are inclined to each other at an angle of 100°, then what is the value of ∠POQ?

#### Answer:

Consider the quadrilateral OPTQ. It is given that ∠PTQ = 100°.

From the property of the tangent we know that the tangent will always be perpendicular to the radius at the point of contact. Therefore we have,

We know that the sum of all angles of a quadrilateral will always be equal to 360°.

Therefore,

+ ++

Let us substitute the values of all the known angles. We have,

Therefore, the value of angle ∠POQ is 80°.

#### Page No 8.55:

#### Question 5:

What the distance between two parallel tangents to a circle of radius 5 cm?

#### Answer:

Two parallel tangents can exist at the two ends of the diameter of the circle. Therefore, the distance between the two parallel tangents will be equal to the diameter of the circle. In the problem the radius of the circle is given as 5 cm. Therefore,

Diameter = 5 × 2

Diameter = 10 cm

Hence, the distance between the two parallel tangents is 10 cm.

#### Page No 8.55:

#### Question 6:

In Q.No. 1, if *PB* = 10 cm, what is the perimeter of Δ *PCD*?

#### Answer:

Here, we have to find the perimeter of triangle PCD.

Perimeter is nothing but sum of all sides of the triangle. Therefore we have,

Perimeter of =

In the given figure we can see that,

=

Therefore,

Perimeter of =

We know that the two tangents drawn to a circle from a common external point will be equal in length. From this property we have,

Now let us replace CQ and QD with CA and DA. We get,

Perimeter of =

Also from the figure we can see that,

Now, let us replace these in the equation for perimeter of. We have,

Perimeter of = PB +PA

Also, from the property of tangents we know that, two tangents drawn to a circle from the same external point will be equal in length. Therefore,

PB = PA

Let us replace PA with PB in the above equation. We get,

Perimeter of = 2PB

It is given in the question that PB = 10 cm. Therefore,

Perimeter of =

Perimeter of = 20 cm

Hence, the perimeter of is 20 cm.

#### Page No 8.55:

#### Question 7:

In the given figure, *CP* and *CQ* are tangents to a circle with centre *O*. *ARB* is another tangent touching the circle at *R*. If *CP* = 11 cm and *BC* = 7 cm, then find the length of *BR*.

#### Answer:

CP and CQ are tangents drawn from an external point C to the circle.

∴ CP = CQ (Length of tangents drawn from an external point to the circle are equal)

⇒ CQ = 11 cm (CP = 11 cm)

BQ = CQ − BC = 11 cm − 7 cm = 4 cm

BR and BQ are tangents drawn from an external point B to the circle.

∴ BR = BQ

⇒ BR = 4 cm ( BQ = 4 cm)

#### Page No 8.55:

#### Question 8:

In the given figure, Δ *ABC* is circumscribing a circle. Find the length of *BC.*

#### Answer:

We are given the following figure

From the figure we get,

BC* = *BP* + *PC* *…… (1)

Now, let us find BP and PC separately.

From the property of tangents we know that when two tangents are drawn to a circle from a common external point, the length of the two tangents from the external point to the respective points of contact will be equal. Therefore we have,

BR = BP

It is given in the problem that BR = 3 cm. Therefore,

BP = 3 cm

Now let us find out PC.

Again using the same property of tangents which says that the length of two tangents drawn to a circle from the same external point will be equal, we have,

PC = QC…… (2)

From the figure we can see that,

QC = AC − AQ…… (3)

Again using the property that length of two tangents drawn to a circle from the same external point will be equal, we have,

AQ = AR

In the problem it is given that,

AR = 4 cm

Therefore,

AQ = 4 cm

Also, the length of* *AC is also given in the problem.

AC = 11 cm

Let us now substitute the values of* *AC and AQ in equation (3)

QC = 11 − 4

QC = 7

From equation (2) we can say that,

PC = 7

Finally, let us substitute the values of PC and BP in equation (1)

BC = BP + PC

BC = 3 + 7

BC = 10

Therefore, length of BC is 10 cm.

#### Page No 8.55:

#### Question 9:

In the given figure, *CP* and *CQ* are tangents from an external point *C* to a circle with centre *O*. *AB* is another tangent which touches the circle at *R*. If *CP* = 11 cm and *BR* = 4 cm, find the length of *BC*.

[

**Hint:**We have,

*CP*= 11 cm

∴

*CP*=

*CQ*=

*CQ*= 11 cm

Now,

*BR*=

*BQ*

⇒

*BQ*= 4 cm

∴

*BC*=

*CQ*−

*BQ*= (11−4)cm = 7 cm

#### Answer:

We are given the following figure

From the figure, we have

BC = CQ − BQ…… (1)

Let us now find out the values of CQ and BQ separately.

From the property of tangents we know that length of two tangents drawn from the same external point will be equal. Therefore,

CP = CQ

It is given in the problem that,

CP = 11

Therefore,

CQ = 11

Now let us find out the value of BQ.

Again from the same property of tangents, we know that length of two tangents drawn from the same external point will be equal. Therefore,

BR = BQ

It is given in the problem that,

BR = 4 cm

Therefore,

BQ = 4 cm

Now let us substitute the values of CQ and BQ in equation (1). We have,

BC = 11 − 4

BC = 7

Therefore, length of BC is 7 cm.

#### Page No 8.56:

#### Question 10:

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

#### Answer:

Let us first draw whatever is given in the problem so that we can understand the problem better.

We have to find the length of AB, which is the chord of the larger circle which touches the smaller circle.

Clearly, OC is the radius of the smaller circle and is touching the tangent AB.

We know that the radius of the circle will always form a right angle with the tangent at the point of contact.

We have draw OA in order to complete the triangle OAC which will be a right triangle.

From the figure it is very clear that OA is the radius of the larger circle which is 5 cm.

We can now find AC using Pythagoras theorem. We have,

Similarly we can find CB. We have,

From the figure we can see that,

AB = AC + CB

Since we have found the values of AC and CB, let us substitute the values in the above equation. We get,

AB = 4 + 4

AB = 8

Therefore, the length of the chord of the larger circle which touches the smaller circle is 8 cm.

#### Page No 8.56:

#### Question 11:

In the given figure, PA and PB are tangents to the circle with centre O such that $\angle \mathrm{APB}=50\xb0$. Write the measure of $\angle \mathrm{OAB}.$ [CBSE 2015]

#### Answer:

It is given that tangents PA and PB are drawn from an external point P to a circle with centre O.

∴ PA = PB (Lengths of tangents drawn from an external point to a circle are equal)

In ∆PAB,

PA = PB

∴ $\angle \mathrm{PBA}=\angle \mathrm{PAB}$ (In a triangle, equal sides have equal angles opposite to them)

Now,

$\angle \mathrm{PAB}+\angle \mathrm{PBA}+\angle \mathrm{APB}=180\xb0$ (Angle sum property)

$\Rightarrow 2\angle \mathrm{PAB}+50\xb0=180\xb0\left(\angle \mathrm{APB}=50\xb0\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 2\angle \mathrm{PAB}=180\xb0-50\xb0=130\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{PAB}=65\xb0\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{PBA}=\angle \mathrm{PAB}=65\xb0.....\left(1\right)$

Now, PA is the tangent and OA is the radius through the point of contact A.

∴ $\angle \mathrm{OAP}=90\xb0$ (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

Now,

$\angle \mathrm{OAB}=\angle \mathrm{OAP}-\angle \mathrm{PAB}=90\xb0-65\xb0=25\xb0\left[\mathrm{Using}\left(1\right)\right]$

Hence, the measure of $\angle \mathrm{OAB}$ is 25º.

#### Page No 8.56:

#### Question 12:

In Fig. 10.85, *PQ* is a chord of a circle and *PT* is the tangent at *P *such that $\angle $*QPT* = 60^{0}. Then , find $\angle $PRQ .

figure

#### Answer:

Construction: Take any point on major arc PQ and name it S. Join SQ and SP.

In the given figure, PT is the tangent. So, PT ⊥ PO.

$\angle \mathrm{QPT}=60\xb0$.

Thus, $\angle \mathrm{OPQ}=90\xb0-60\xb0=30\xb0$

OQ = OP (Radii of the circle)

$\angle \mathrm{OQP}=\angle \mathrm{OPQ}=30\xb0$

$\mathrm{In}\u25b3\mathrm{OPQ},\phantom{\rule{0ex}{0ex}}\angle \mathrm{OPQ}+\angle \mathrm{OQP}+\angle \mathrm{POQ}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 30\xb0+30\xb0+\angle \mathrm{POQ}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{POQ}=120\xb0$

Now, $\angle \mathrm{PSQ}=\frac{1}{2}\angle \mathrm{POQ}=\frac{1}{2}\times 120=60\xb0\phantom{\rule{0ex}{0ex}}$

PSQR is a cyclic quadrilateral. Thus,

$\angle \mathrm{PSQ}+\angle \mathrm{PRQ}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 60\xb0+\angle \mathrm{PRQ}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{PRQ}=120\xb0$

#### Page No 8.56:

#### Question 13:

In Fig. 10.86, *PQL* and *PRM *are tangents to the circle with centre *O* at the points *Q* and* R* respectively and *S *is a point on the circle such that * $\angle $SQL* = 50^{0 }and* $\angle $SRM* = 60^{0}. Then , find $\angle $*QSR*.

figure

#### Answer:

In the given figure,

PL and PM are the tangents to the circle with centre O.

$\angle \mathrm{SQL}=50\xb0,\angle \mathrm{SRM}=60\xb0$

$\mathrm{PL}\perp \mathrm{OQ}\mathrm{and}\mathrm{PM}\perp \mathrm{OR}$

Thus,

$\angle \mathrm{ORS}=90\xb0-60\xb0=30\xb0\phantom{\rule{0ex}{0ex}}\angle \mathrm{OQS}=90\xb0-50\xb0=40\xb0$

OQ = OS

So, $\angle \mathrm{OQS}=\angle \mathrm{OSQ}=40\xb0$

Similarly.

OS = OR

So, $\angle \mathrm{ORS}=\angle \mathrm{OSR}=30\xb0$

$\angle \mathrm{RSQ}=\angle \mathrm{QSO}+\angle \mathrm{RSO}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{RSQ}=40\xb0+30\xb0=70\xb0$

#### Page No 8.56:

#### Question 14:

In Fig. 10.87 , *BOA* is a diameter of a circle and the tangent at a point *P* meets *BA *produced at* T* . If $\angle $*PBO* = 30^{0} , then find $\angle $*PTA* .

figure

#### Answer:

In the given figure, PT is the tangent to the circle with centre O.

$\angle $PBO = 30$\xb0$

OP = OB (Radii)

So, $\angle $PBO = $\angle $BPO = 30$\xb0$

$\angle $BPA = 90$\xb0$ (Angle made by the diameter on the arc of the circle)

In ∆APB,

$\angle $BPA + $\angle $PBO + $\angle $PAB = 180$\xb0$

⇒ 90$\xb0$ + 30$\xb0$ + $\angle $PAB = 180$\xb0$

⇒ $\angle $PAB = 60$\xb0$

In ∆OPA,

OA = OP (Radii)

So, $\angle $OPA = OAP = 60$\xb0$

Hence, $\angle $AOP = 180$\xb0$ − 60$\xb0$ − 60$\xb0$ = 60$\xb0$

In ∆OPT,

$\angle $OPT + $\angle $PTO + $\angle $POT = 180$\xb0$

⇒ 90$\xb0$ + $\angle $PTO + 60$\xb0$ = 180$\xb0$

⇒ $\angle $PTO = 30$\xb0$

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