**NCERT Solutions for Class 10 Maths Chapter 10 Circles. Question Answer.**

Page No: 209

**Exercise: 10.1**

1. How many tangents can a circle have?

**Answer**

A circle can have infinite tangents.

2. Fill in the blanks :

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

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

(iii) A circle can have ............... parallel tangents at the
most.

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

**Answer**

(i) one

(ii) secant

(iii) two

(iv) point of contact

3. 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 so that OQ = 12 cm. Length PQ is :

a point Q so that OQ = 12 cm. Length PQ is :

(A) 12 cm

(B) 13 cm

(C) 8.5 cm

(D) √119 cm

(B) 13 cm

(C) 8.5 cm

(D) √119 cm

**Answer**

∴ OP ⊥ PQ

By Pythagoras theorem in ΔOPQ,

OQ

⇒ (12)

^{2}= OP^{2}+^{ }PQ^{2}⇒ (12)

^{2 }= 5^{2}+ PQ^{2}
⇒PQ

^{2}= 144 - 25
⇒PQ

^{2}= 119
⇒PQ = √119 cm

(D) is the correct option.

4. Draw a circle and two lines parallel to a given line such that one is a
tangent and the

other, a secant to the circle.

**Answer**

AB and XY are two parallel lines where AB is the tangent to the circle at
point C while XY is the secant to the circle.

Page NO: 213

**Exercise: 10.2**

In Q.1 to 3, choose the correct option and give justification.

1. 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

(B) 12 cm

(C) 15 cm

(D) 24.5 cm

**Answer**

The line drawn from the centre of the circle to the tangent is perpendicular
to the tangent.

∴ OP ⊥ PQ

also, ΔOPQ is right angled.

OQ = 25 cm and PQ = 24 cm (Given)

also, ΔOPQ is right angled.

OQ = 25 cm and PQ = 24 cm (Given)

By Pythagoras theorem in ΔOPQ,

OQ

⇒ (25)

^{2}= OP^{2}+^{ }PQ^{2}⇒ (25)

^{2 }= OP^{2}+ (24)^{2}
⇒ OP

^{2}= 625 - 576
⇒ OP

^{2}= 49
⇒ OP = 7 cm

The radius of the circle is option (A) 7 cm.

2. In Fig. 10.11, if TP and TQ are the 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°

2. In Fig. 10.11, if TP and TQ are the 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**

OP and OQ are radii of the circle to the tangents TP and TQ respectively.

∴ OP ⊥ TP and,

∴ OQ ⊥ TQ

∠OPT = ∠OQT = 90°

In quadrilateral POQT,

Sum of all interior angles = 360°

∠PTQ + ∠OPT + ∠POQ + ∠OQT = 360°

⇒ ∠PTQ + 90° + 110° + 90° = 360°

⇒ ∠PTQ = 70°

∠PTQ is equal to option (B) 70°.

3. If tangents PA and PB from a point P to a circle with centre O are
inclined to each other at angle of 80°, then ∠ POA is equal to

(A) 50°

(B) 60°

(C) 70°

(D) 80°

(A) 50°

(B) 60°

(C) 70°

(D) 80°

**Answer**

OA and OB are radii of the circle to the tangents PA and PB respectively.

∴ OA ⊥ PA and,

∴ OB ⊥ PB

∠OBP = ∠OAP = 90°

∴ OA ⊥ PA and,

∴ OB ⊥ PB

∠OBP = ∠OAP = 90°

In quadrilateral AOBP,

Sum of all interior angles = 360°

∠AOB + ∠OBP + ∠OAP + ∠APB = 360°

⇒ ∠AOB + 90° + 90° + 80° = 360°

Sum of all interior angles = 360°

∠AOB + ∠OBP + ∠OAP + ∠APB = 360°

⇒ ∠AOB + 90° + 90° + 80° = 360°

⇒ ∠AOB = 100°

Now,

In ΔOPB and ΔOPA,AP = BP (Tangents from a point are equal)

OA = OB (Radii of the circle)

OP = OP (Common side)

∴ ΔOPB ≅ ΔOPA (by SSS congruence condition)

Thus ∠POB = ∠POA

∠AOB = ∠POB + ∠POA

⇒ 2 ∠POA = ∠AOB

⇒ ∠POA = 100°/2 = 50°

∠POA is equal to option (A) 50°

Page No: 214

4. Prove that the tangents drawn at the ends of a diameter of a
circle are parallel.

**Answer**

∴ OB ⊥ RS and,

∴ OA ⊥ PQ

∠OBR = ∠OBS = ∠OAP = ∠OAQ = 90º

∠OBR = ∠OBS = ∠OAP = ∠OAQ = 90º

From the figure,

∠OBR = ∠OAQ (Alternate interior angles)

∠OBS = ∠OAP (Alternate interior angles)

Since alternate interior angles are equal, lines PQ and RS will be parallel.

∠OBR = ∠OAQ (Alternate interior angles)

∠OBS = ∠OAP (Alternate interior angles)

Since alternate interior angles are equal, lines PQ and RS will be parallel.

Hence Proved that the tangents drawn at the ends of a diameter of a circle are
parallel.

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

**Answer**

Let AB be the tangent to the circle at point P with centre O.

We have to prove that PQ passes through the point O.

Suppose that PQ doesn't passes through point O. Join OP.

Through O, draw a straight line CD parallel to the tangent AB.

PQ intersect CD at R and also intersect AB at P.

AS, CD // AB PQ is the line of intersection,

∠ORP = ∠RPA (Alternate interior angles)
but also,

∠RPA = 90° (PQ ⊥ AB)

⇒ ∠ORP = 90°

∠ROP + ∠OPA = 180° (Co-interior angles)

⇒∠ROP + 90° = 180°

⇒∠ROP = 90°

Thus, the ΔORP has 2 right angles i.e. ∠ORP and ∠ROP which is not
possible.

Hence, our supposition is wrong.

∴ PQ passes through the point O.

6. The length of a tangent from a point A at distance 5 cm from the
centre of the circle is 4 cm. Find the radius of the circle.

**Answer**

OA = 5cm and AB = 4 cm (Given)

In ΔABO,

By Pythagoras theorem in ΔABO,

OA

^{2}=

^{ }AB

^{2 }+ BO

^{2}

⇒ 5

^{2 }= 4

^{2 }+ BO

^{2}

⇒ BO

^{2}= 25 - 16

⇒ BO

^{2}= 9

⇒ BO = 3

∴ The radius of the circle is 3 cm.

7. 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**

AB be the chord of the larger circle which touches the smaller circle at point
P.

∴ AB is tangent to the smaller circle to the point P.

⇒ OP ⊥ AB

By Pythagoras theorem in ΔOPA,

OA

⇒ 5

⇒ AP

⇒ AP = 4

In ΔOPB,

Since OP ⊥ AB,

AP = PB (Perpendicular from the center of the circle bisects the chord)

AB = 2AP = 2 × 4 = 8 cm

∴ The length of the chord of the larger circle is 8 cm.

8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC

From the figure we observe that,

DR = DS (Tangents on the circle from point D) … (i)

AP = AS (Tangents on the circle from point A) … (ii)

By Pythagoras theorem in ΔOPA,

OA

^{2}= AP^{2}+ OP^{2}⇒ 5

^{2}= AP^{2}+ 3^{2}⇒ AP

^{2}= 25 - 9⇒ AP = 4

In ΔOPB,

Since OP ⊥ AB,

AP = PB (Perpendicular from the center of the circle bisects the chord)

AB = 2AP = 2 × 4 = 8 cm

∴ The length of the chord of the larger circle is 8 cm.

8. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC

**Answer**

From the figure we observe that,

DR = DS (Tangents on the circle from point D) … (i)

AP = AS (Tangents on the circle from point A) … (ii)

BP = BQ (Tangents on the circle from point B) … (iii)

CR = CQ (Tangents on the circle from point C) … (iv)

Adding all these equations,

DR + AP + BP + CR = DS + AS + BQ + CQ

⇒ (BP + AP) + (DR + CR) = (DS + AS) + (CQ + BQ)

CR = CQ (Tangents on the circle from point C) … (iv)

Adding all these equations,

DR + AP + BP + CR = DS + AS + BQ + CQ

⇒ (BP + AP) + (DR + CR) = (DS + AS) + (CQ + BQ)

⇒ CD + AB = AD + BC

9. In Fig. 10.13, XY and X′Y′ are two parallel tangents to a circle with
centre O and another tangent AB with point of contact C intersecting XY at A
and X′Y′ at B. Prove that ∠ AOB = 90°.

**Answer**

We joined O and C

In ΔOPA and ΔOCA,

OP = OC (Radii of the same circle)

AP = AC (Tangents from point A)

AO = AO (Common side)

∴ ΔOPA ≅ ΔOCA (SSS congruence criterion)

⇒ ∠POA = ∠COA … (i)

Similarly,

ΔOQB ≅ ΔOCB

∠QOB = ∠COB … (ii)

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

∴ ∠POA + ∠COA + ∠COB + ∠QOB = 180 º

From equations (i) and (ii),

2∠COA + 2∠COB = 180º

⇒ ∠COA + ∠COB = 90º

⇒ ∠AOB = 90°

∠QOB = ∠COB … (ii)

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

∴ ∠POA + ∠COA + ∠COB + ∠QOB = 180 º

From equations (i) and (ii),

2∠COA + 2∠COB = 180º

⇒ ∠COA + ∠COB = 90º

⇒ ∠AOB = 90°

10. Prove that the angle between the two tangents drawn from an
external point to a circle is supplementary to the angle subtended by the
line-segment joining the points of contact at the centre.

**Answer**

It can be observed that

OA ⊥ PA

∴ ∠OAP = 90°

Similarly, OB ⊥ PB

∴ ∠OBP = 90°

In quadrilateral OAPB,

Sum of all interior angles = 360º

∠OAP +∠APB +∠PBO +∠BOA = 360º

⇒ 90º + ∠APB + 90º + ∠BOA = 360º

⇒ ∠APB + ∠BOA = 180º

∴ The angle between the two tangents drawn from an external point to a circle
is supplementary to the angle subtended by the line-segment joining the points
of contact at the centre.

ABCD is a parallelogram,

∴ AB = CD ... (i)

∴ BC = AD ... (ii)

CR = CQ (Tangents to the circle at C)

BP = BQ (Tangents to the circle at B)

AP = AS (Tangents to the circle at A)

Adding all these,

DR + CR + BP + AP = DS + CQ + BQ + AS

⇒ (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)

In ΔABC,

11. Prove that the parallelogram circumscribing a circle is a
rhombus.

**Answer**

∴ AB = CD ... (i)

∴ BC = AD ... (ii)

From the figure, we observe that,

DR = DS (Tangents to the circle at D)CR = CQ (Tangents to the circle at C)

BP = BQ (Tangents to the circle at B)

AP = AS (Tangents to the circle at A)

Adding all these,

DR + CR + BP + AP = DS + CQ + BQ + AS

⇒ (DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)

⇒ CD + AB = AD + BC ... (iii)

Putting the value of (i) and (ii) in equation (iii) we get,

2AB = 2BC

⇒ AB = BC ... (iv)

By Comparing equations (i), (ii), and (iv) we get,

AB = BC = CD = DA

∴ ABCD is a rhombus.

Putting the value of (i) and (ii) in equation (iii) we get,

2AB = 2BC

⇒ AB = BC ... (iv)

By Comparing equations (i), (ii), and (iv) we get,

AB = BC = CD = DA

∴ ABCD is a rhombus.

12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such
that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm
and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

**Answer**

Length of two tangents drawn from the same point to the circle are equal,

∴ CF = CD = 6cm

∴ BE = BD = 8cm

∴ AE = AF =

= √(14 + ∴ CF = CD = 6cm

∴ BE = BD = 8cm

∴ AE = AF =

*x*
We observed that,

AB = AE + EB =

BC = BD + DC = 8 + 6 = 14

CA = CF + FA = 6 +

AB = AE + EB =

*x*+ 8BC = BD + DC = 8 + 6 = 14

CA = CF + FA = 6 +

*x*
Now semi perimeter of triangle (s) is,

⇒ 2s = AB + BC + CA

=

= 28 + 2

⇒s = 14 +

⇒ 2s = AB + BC + CA

=

*x*+ 8 + 14 + 6 +*x*= 28 + 2

*x*⇒s = 14 +

*x*
Area of ΔABC = √s (s - a)(s - b)(s - c)

*x*) (14 +

*x*-

*14)(14 +*

*x*-

*x*- 6)(14 +

*x*-

*x -*8)

= √(14 +

= √(14 + *x*) (*x*)(8)(6)*x*) 48

*x*... (i)

also, Area of ΔABC = 2×area of (ΔAOF + ΔCOD + ΔDOB)

= 2×[(1/2×OF×AF) + (1/2×CD×OD) + (1/2×DB×OD)]

= 2×1/2 (4

*x*+ 24 + 32) = 56 + 4*x*... (ii)
Equating equation (i) and (ii) we get,

√(14 +

Squaring both sides,*x*) 48*x*= 56 + 4*x*
48

*x*(14 +*x*) = (56 + 4*x*)^{2}
⇒ 48

*x =*[4(14 + x)]^{2}/(14 +*x*)
⇒ 48

*x =*16 (14 +*x*)
⇒ 48

*x =*224 + 16*x*
⇒ 32

*x =*224
⇒

Hence, AB = *x =*7 cm*x*+ 8 = 7 + 8 = 15 cm

CA = 6 +

*x*= 6 + 7 = 13 cm

13. Prove that opposite sides of a quadrilateral circumscribing a
circle subtend supplementary angles at the centre of the circle.

**Answer**

In ΔOAP and ΔOAS,

AP = AS (Tangents from the same point)

OP = OS (Radii of the circle)

OA = OA (Common side)

ΔOAP ≅ ΔOAS (SSS congruence condition)

∴ ∠POA = ∠AOS

⇒∠1 = ∠8

Similarly we get,

∠2 = ∠3

∠4 = ∠5

∠6 = ∠7

Similarly we get,

∠2 = ∠3

∠4 = ∠5

∠6 = ∠7

Adding all these angles,

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 +∠8 = 360º

⇒ (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º

⇒ 2 ∠1 + 2 ∠2 + 2 ∠5 + 2 ∠6 = 360º

⇒ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º

⇒ (∠1 + ∠2) + (∠5 + ∠6) = 180º

⇒ ∠AOB + ∠COD = 180º

Similarly, we can prove that ∠ BOC + ∠ DOA = 180º

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 +∠8 = 360º

⇒ (∠1 + ∠8) + (∠2 + ∠3) + (∠4 + ∠5) + (∠6 + ∠7) = 360º

⇒ 2 ∠1 + 2 ∠2 + 2 ∠5 + 2 ∠6 = 360º

⇒ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º

⇒ (∠1 + ∠2) + (∠5 + ∠6) = 180º

⇒ ∠AOB + ∠COD = 180º

Similarly, we can prove that ∠ BOC + ∠ DOA = 180º

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.