## RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

**Other Exercises**

- RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.1
- RD Sharma Class 10 Solutions Chapter 8 Circles Ex 8.2
- RD Sharma Class 10 Solutions Chapter 8 Circles VSAQS
- RD Sharma Class 10 Solutions Chapter 8 Circles MCQS

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

**Question 1.**

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) √119 cm

**Solution:**

**(d)** Radius of a circle OP = 5 cm OQ = 12 cm, PQ is tangent

OP ⊥ PQ

In right ∆OPQ,

OQ² = OP² + PQ² (Pythagoras Theorem)

=> (12)² = (5)2 + PQ²

=> 144 = 25 + PQ²

PQ² = 144 – 25 = 119

PQ = √119

**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

**Solution:**

**(a)** Let PQ be the tangent from Q to the circle with O as centre

PQ = 24 cm

OQ = 25 cm

Let Radius OQ = r

OQ ⊥ PQ

Now in right ∆OPQ,

OQ² = OP² + PQ² (Pythagoras Theorem)

=> (25)² = r² + (24)²

=> 625 = r² + 576

=> r² = 625 – 576 = 49 = (7)²

r = 7

Radius of the circle = 7 cm

**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) √7 cm

(b) 7 cm

(c) 5 cm

(d) 25 cm

**Solution:**

**(c)** Let AB be the tangent from A to the circle of centre O, then

OB = 3 cm

BA = 4 cm

OB ⊥ BA

In right ∆OBA,

OA² = OB² + BA² (Pythagoras Theorem) = (3)² + (4)² = 9 + 16 = 25 = (5)²

OA = 5

Distance of A from the centre O = 5 cm

**Question 4.**

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

(a) 50°

(b) 60°

(c) 70°

(d) 80°

**Solution:**

**(a)** PA and PB are the tangents to the circle from P and ∠APB = 80°

∠AOB = 180° – ∠APB = 180°- 80° = 100°

But OP is the bisector of ∠AOB

∠POA = ∠POB = ∠AOB

=> ∠POA = x 100° = 50°

**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°

**Solution:**

**(b)** TP and TQ are the tangents from T to the circle with centre O and OP, OQ are joined and ∠POQ = 110°

But ∠POQ + ∠PTQ = 180°

=> 110° + ∠PTQ = 180°

=> ∠PTQ = 180° – 110° = 70°

**Question 6.**

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

(a) 30°

(b) 45°

(c) 60°

(d) 90°

**Solution:**

**(b)** In a circle with centre O, PQ is a tangent to the circle at P and ∆OPQ is an isosceles triangle such that OP = PQ

OP is radius of the circle

OP ⊥ PQ

OP = PQ

∠POQ = ∠OQP

But ∠POQ = ∠PQO = 90° (OP ⊥ PQ)

∠OQP = ∠POQ = 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°

**Solution:**

**(d)** Two circles with centres O and O’ touch each other at C externally

A common tangent is drawn which touches the circles at A and B respectively.

Join OA, O’B and O’O which passes through C

AO = BO’ (radii of the equal circle)

AB || OO’

=> AOO’B is a rectangle

Draw another common tangent through C which intersects AB at D, then DA = DC = DB

ADCO and BDCO’ are squares

AC and BC are the diagonals of equal square

AC = BC

∠DAC = ∠DBC = 45°

∠ACB = 90°

**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 ∆ABC. The radius of the circle is

(a) 1 cm

(b) 2 cm

(c) 3 cm

(d) 4 cm

**Solution:**

**(b)** In a right ∆ABC, ∠B = 90°

BC = 6 cm, AB = 8 cm

AC² = AB² + BC² (Pythagoras Theorem) = (8)² + (6)² = 64 + 36 = 100 = (10)²

AC = 10 cm

An incircle is drawn with centre 0 which touches the sides of the triangle ABC at P, Q and R

OP, OQ and OR are radii and AB, BC and CA are the tangents to the circle

OP ⊥ AB, OQ ⊥ BC and OR ⊥ CA

OPBQ is a square

Let r be the radius of the incircle

PB = BQ = r

AR = AP = 8 – r,

CQ = CR = 6 – r

AC = AR + CR

=> 10 = 8 – r + 6 – r

10 = 14 – 2r

=> 2r = 14 – 10 = 4

=> r = 2

Radius of the incircle = 2 cm

**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°

**Solution:**

**(c)** PQ is a tangent to the circle with centre O, from P, QOR is the diameter and ∠POR = 120°

OQ is radius and PQ is tangent to the circle

OQ ⊥ QP or ∠OQP = 90°

But ∠QOP + ∠POR = 180° (Linear pair)

=> ∠QOP + 120° = 180°

∠QOP = 180° – 120° = 60°

Now in ∆POQ

∠QOP + ∠OQP + ∠OPQ = 180° (Angles of a triangle)

=> 60° + 90° + ∠OPQ = 180°

=> 150° + ∠OPQ = 180°

=> ∠OPQ = 180° – 150° = 30°

**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

**Solution:**

**(b)** A circle is inscribed in a quadrilateral ABCD which touches the sides AB, BC, CD and DA at P, Q, R and S respectively then the sum of two opposite sides is equal to the sum of other two opposite sides

AB + CD = BC + AD

**Question 11.**

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

(a) √7 cm

(b) 2√7cm

(c) 10 cm

(d) 5 cm

**Solution:**

**(b)** Radius of the circle = 6 cm

and distance of the external point from the centre = 8 cm

Length of tangent = √{(8)² – (6)²}

= √(64 – 36) = √28

= √(4 x 7) = 2√7 cm

**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

**Solution:**

**(c)** AB and CD are two common tangents to the two circles which touch each other externally at C and intersect AB in D

CD = 4 cm

DA and DC are tangents to the first circle from D

CD = AD = 4 cm

Similarly DC and DB are tangents to the second circle from D

CD = DB = 4 cm

AB = AD + DB = 4 + 4 = 8 cm

**Question 13.**

In the adjoining 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

**Solution:**

**(b)** AD, AE and BC are the tangents to the circle at D, E and F respectively

D and AE are tangents to the circle from A

AD = AE ……(i)

Similarly, CD = CF and BE = BF ….(ii)

Now AB + AC + BC = AE – BE + AD – CD + CF + BF

= AD – BE + AD – CD + BE + BE

= 2AD [From (i) and (ii)]

or 2 AD = AB + BC + CA

**Question 14.**

In the 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

**Solution:**

**(d)** In the figure, 0 is the centre of the circle

QR is tangent to the circle and QOS is a diameter SQ = 6 cm, QR = 4 cm

OQ = QS = x 6 = 3 cm

OQ is radius

OQ ⊥ QR

Now in right ∆OQR

OR² = QR² + QO² = (3)² + (4)² = 9 + 16 = 25 = (5)²

OR = 5 cm

**Question 15.**

In the figure, the perimeter of ∆ABC is

(a) 30 cm

(b) 60 cm

(c) 45 cm

(d) 15 cm

**Solution:**

**(a)** ∆ABC is circumscribed of circle with centre O

AQ = 4 cm, CP = 5 cm and BR = 6 cm

AQ and AR the tangents to the circle AQ = AR = 4 cm

Similarly BP and BR are tangents,

BP = BR = 6 cm

and CP and CQ are the tangents

CQ = CP = 5 cm

AB = AR + BR = 4 + 6 = 10 cm

BC = BP + CP = 6 + 5 = 11 cm

AC = AQ + CQ = 4 + 5 = 9 cm

Perimeter of ∆ABC = AB + BC + AC = 10 + 11 + 9 = 30 cm

**Question 16.**

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

(a) 2√2 cm

(b) 2 cm

(c) 2√3 cm

(d) 3√2 cm

**Solution:**

**(c)** In the figure, AP is the tangent to the circle with centre O such that

OP = 4 cm, ∠OPA = 30°

Join OA, let AP = x

**Question 17.**

AP and AQ 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

**Solution:**

**(c)** OP is radius, PA is the tangent

OP ⊥ AP

Now in right ∆OAP,

OA² = OP² + AP²

(15)² = (9)² + AP²

225 = 81 + AP²

=> AP² = 225 – 81 = 144 = (12)²

AP = 12 cm

But AP = AQ = 12 cm (tangents from A to the circle)

AP + AQ = 12+ 12 = 24 cm

**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

**Solution:**

**(d)** In the figure, PQ is diameter XPY is tangent to the circle with centre O and radius 5 cm

From P, at a distance of 8 cm AB is a chord drawn parallel to XY

To find the length of AB

Join OA

XY is tangent and OP is the radius

OP ⊥ XY or PQ ⊥ XY

AB || XY

OQ is ⊥ AB which meets AB at R

Now in right ∆OAR,

OA² = OR² + AR²

(5)² = (3)² + AR²

25 = 9 + AR²

=> AR² = 25 – 9 = 16 = (4)²

AR = 4 cm

But R is mid-point of AB

AB = 2 AR = 2 x 4 = 8 cm

**Question 19.**

If PT is tahgent drawn froth 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°

**Solution:**

**(c)** In the figure, PT is the tangent to the circle with centre O.

OP and OT are joined

PT is tangent and OT is the radius

OT ⊥ PT

Now in right ∆OPT

∠OTP = 90°

∠OPT + ∠POT = 180° – 90° = 90°

**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

**Solution:**

**(d)** In the figure, ∆ABC is the circumscribed a circle

AB = 12 cm, BC = 8 cm and AC = 10 cm

Let AD = a, DB = b and EC = c, then

AF = a, BE = b and FC = c

But AB + BC + AC = 12 + 8 + 10 = 30

a + b + b + c + c + a = 30

=> 2 (a + b + c) = 30

a + b + c = 15

Subtracting BC or b + c from this a = 15 – 8 = 7

AD = 7 cm

**Question 21.**

In the figure, if AP = PB, then

(a) AC = AB

(b) AC = BC

(c) AQ = QC

(d) AB = BC

**Solution:**

**(b)** In the figure, AP = PB

But AP and AQ are the tangent from A to the circle

AP = AQ

Similarly PB = BR

But AP = PB (given)

AQ = BR ….(i)

But CQ and CR the tangents drawn from C to the circle

CQ = CR

Adding in (i)

AQ + CQ = BR + CR

AC = BC

**Question 22.**

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

(a) √91 cm

(b) √127 cm

(c) √119 cm

(d) √109 cm

**Solution:**

**(b)** In the figure,

OA = 6 cm, OB = 3 cm and AP = 10 cm

OA is radius and AP is the tangent

OA ⊥ AP

Now in right ∆OAP

OP² = AP² + OA² = (10)² + (6)² = 100 + 36 = 136

Similarly BP is tangent and OB is radius

OP² = OB² + BP²

136 = (3)² + BP2

=> 136 = 9 + BP²

=> BP² = 136 – 9 = 127

BP = √127 cm

**Question 23.**

In the 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°

**Solution:**

**(c)** In the figure, PR is the tangent to the circle at P.

O is the centre of the circle ∠QPR = 60°

OP is the radius and PR is the tangent OPR = 90°

=> ∠OPQ + ∠QPR = 90°

=> ∠OPQ + 60° = 90°

=> ∠OPQ = 90° – 60° = 30°

OP = OQ (radii of the circle)

∠OQP = 30°

In ∆OPQ,

∠OPQ + ∠OQP + ∠POQ = 180°

=> 30° + 30° + ∠PQR = 180°

=> 60° + ∠POQ = 180°

∠POQ = 180° – 60° = 120°

**Question 24.**

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

(a) PQ

(b) QR

(c) PR

(d) PS

**Solution:**

**(a)** In the figure, quadrilateral PQRS is circumscribed a circle

PD = PA (tangents from P to the circle)

Similarly QA = QB

PD + QB = PA + QA = PQ

**Question 25.**

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

(a) 9 cm

(b) 18 cm

(c) 15 cm

(d) 13.5 cm

**Solution:**

**(a)** In the figure, two equal circles touch, each other externally at T

QR is the common tangent

QP = 4.5 cm

PQ = PT (tangents from P to the circle)

Similarly PT = PR

PQ = PT = PR

Now QR = PQ + PR = 4.5 + 4.5 = 9 cm

**Question 26.**

In the 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°

**Solution:**

**(a)** In the figure, APB is a tangent to the circle with centre O

∠QPB = 50°

OP is radius and APB is a tangent

OP ⊥ AB

=> ∠OPB = 90°

=> ∠OPQ + ∠QPB = 90°

∠OPQ + 50° = 90°

=> ∠OPQ = 90° – 50° = 40°

But OP = OQ

∠OPQ = OQP = 40°

∠POQ = 180°- (40° + 40°) = 180° – 80° = 100°

**Question 27.**

In the 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

**Solution:**

**(c)** In the figure, PA and PB are the tangents to the circle with centre O

∠APB = 30°

Chord AC || BP,

AB is joined

PA = PB

∠PAB = ∠PBA

But ∠PAB + ∠PBA = 180° – 30° = 150°

=> ∠BPA + ∠PBA = 150°

=> 2 ∠PBA = 150°

=> ∠PBA = 75°

AC || BC

∠BAC = ∠PBA = 75°

But ∠PBA = ∠ACB = 75° (Angles in the alternate segment)

∠ABC = 180° – (75° + 75°) = 180° – 150° = 30°

**Question 28.**

In the figure, PR =

(a) 20 cm

(b) 26 cm

(c) 24 cm

(d) 28 cm

**Solution:**

**(b)** In the figure, two circles with centre O and O’ touch each other externally

PQ and RS are the tangents drawn to the circles

OQ and O’S are the radii of these circles and

OQ = 3 cm, PQ = 4 cm O’S = 5 cm and SR = 12 cm

Now in right ∆OQP

OP² = (OQ)² + PQ² = (3)² + (4)² = 9 + 16 = 25 = (5)²

OP = 5 cm

Similarly in right ∆RSO’

(O’R)² = (RS)² + (O’S)² = (12)² + (5)² = 144 + 25 = 169 = (13)²

O’R = 13 cm

Now PR = OP + OO’ + O’R = 5 + (3 + 5) + 13 = 26 cm

**Question 29.**

Two circles of same radii r and centres O and O’ touch each other at P as shown in figure. If OO’ is produced to meet the circle C (O’, r) at A and AT is a tangent to the circle C (O, r) such that O’Q ⊥ AT. Then AO : AO’ =

(a)

(b) 2

(c) 3

(d)

**Solution:**

**(c)** Two circles of equal radii touch each other externally at P. OO’ produced meets at A

From A, AT is the tangent to the circle (O, r)

O’Q ⊥ AT

Now AO : AO’ = 3r : r

= 3 : 1 =

**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

**Solution:**

**(c)** In the figure, two concentric circles of radii 3 cm and 5 cm with centre O

Chord BC touches the inner circle at P

Draw a tangent AB to the inner circle

Join OQ and OA

OQ is radius and AQB is the tangent

OQ ⊥ AB and OQ bisects AB

AQ = QB

Similarly, BP = PC or P is mid-point of BC

But BQ and BP are tangents from B

QB = BP = AQ

In right ∆OAQ,

OA² = AQ² + OQ²

(5)² = AQ² + (3)²

=> AQ² = (5)² – (3)²

=> AQ² = 25 – 9 = 16 = (4)²

AQ = 4 cm

BC = 2 BP = 2 BQ = 2 AQ = 2 x 4 = 8 cm

**Question 31.**

In the 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

**Solution:**

**(c)** In the figure, two concentric circles with centre O

From a point P on the outer circle,

PRT and PQS are the tangents are drawn to the inner circle at R and Q respectively

PR = 7.5 cm

Join OR and OQ

PT is chord and OR is radius

R is mid-point of PT

Similarly Q is mid-point of PS

But PR = PQ (tangents from P)

PT = 2 PR and PS = 2 PQ

PS = 2 PQ = 2 PR = 2 x 7.5 = 15 cm

**Question 32.**

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

(a) 11 cm

(b) 7 cm

(c) 5 cm

(d) 3 cm

**Solution:**

**(c)** In the figure, AB and AC are the tangents to the circle from A

DE is another tangent drawn from P

AB = 8 cm, PE = 3 cm

AB = AC (tangents drawn from A to the circle)

Similarly PE = EC and DP = DB

Now AE = AC – CE = AB – PE = 8 – 3 = 5 cm

**Question 33.**

In the 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°

**Solution:**

**(b)** In the figure, PQ and PR are the tangents drawn from P to the circle with centre O

∠OPQ = 35°

PO is joined

PQ = PR (tangents from P to the circle)

∠OPQ = ∠OPR

=> 35° = a

=> a = 35°

OQ is radius and PQ is tangent

OQ ⊥ PQ

=> ∠OQP = 90°

In ∆OQP,

∠POQ + ∠QPO = 90°

=> b + 35° = 90°

=> b = 90° – 35° = 55°

a = 35°, b = 55°

**Question 34.**

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

(a) 25°

(b) 30°

(c) 40°

(d) 60°

**Solution:**

**(b)** In the figure, TP and TQ are the tangents drawn from T to the circle with centre O

OP, OQ and PQ are joined

∠TQP = 60°

TP = TQ (Tangents from T to the circle)

∠TQP = ∠TPQ = 60°

∠PTQ = 180° – (60° + 60°) = 180° – 120° = 60°

and ∠POQ = 180° – ∠PTQ = 180° – 60° = 120°

But OP = OQ (radii of the same circle)

∠OPQ = ∠OQP

But ∠OPQ + ∠OQP = 180° – 120° = 60°

But ∠OPQ = 30°

**Question 35.**

In the 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 **[CBSE 2012]**

(a) 11 cm

(b) 10 cm

(c) 14 cm

(d) 15 cm

**Solution:**

**(b)** In the figure,

PA = 4 cm, BP = 3 cm, AC = 11 cm

AP and AR are the tangents from A to the circle

AP = AR

=> AR = 4 cm

Similarly BP and BQ are tangents

BQ = BP = 3 cm

AC =11 cm

AR + CR = 11 cm

4 + CR =11 cm

CR = 11 – 4 = 7 cm

CQ and CR are tangents to the circle

CQ = CR = 7 cm

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

**Question 36.**

In the figure, a circle touches the side DF of AEDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm, then the perimeter of ∆EDF is **[CBSE 2012]**

(a) 18 cm

(b) 13.5 cm

(c) 12 cm

(d) 9 cm

**Solution:**

**(a)** In ∆DEF

DF touches the circle at H

and circle touches ED and EF Produced at K and M respectively

EK = 9 cm

EK and EM are the tangents to the circle

EM = EK = 9 cm

Similarly DH and DK are the tangent

DH = DK and FH and FM are tangents

FH = FM

Now, perimeter of ∆DEF

= ED + DF + EF

= ED + DH + FH + EF

= ED + DK + EM + EF

= EK + EM

= 9 + 9

= 18 cm

**Question 37.**

In the 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 **[CBSE 2013]**

(a) 3 cm

(b) 5 cm

(c) 4 cm

(d) 6 cm

**Solution:**

**(b)** If figure, DE and DF are tangents to the circle drawn from D.

A is the centre of the circle.

DE = 5 cm and DE ⊥ DF

Join AE, AF

DE is the tangent and AE is radius

AE ⊥ DE

Similarly, AF ⊥ DF

But ∠D = 90° (given)

AFDE is a square

AE = DE (side of square)

But DE = 5 cm

AE = 5 cm

Radius of circle is 5 cm

**Question 38.**

In the 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 **[CBSE 2013]**

(a) 11

(b) 18

(c) 6

(d) 15

**Solution:**

**(a)** In the figure, a circle touches the sides of a quadrilateral ABCD

∠B = 90°, OP = OQ = r

AB = 29 cm, AD = 23 cm, DS = 5 cm

∠B = 90°

BA is tangent and OQ is radius

∠OQB = 90°

Similarly OP is radius and BC is tangents

∠OPB = 90°

But ∠B = 90° (given)

PBQO is a square

DS = 5 cm

But DS and DR are tangents to the circles

DR = 5 cm

But AD = 23 cm

AR = 23 – 5= 18 cm

AR = AQ (tangents to the circle from A)

AQ = 18 cm

But AB = 29 cm

BQ = 29 – 18 = 11 cm

OPBQ is a square

OQ = BQ = 11 cm

Radius of the circle = 11 cm

**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

**Solution:**

**(c)**

**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 ∠APB is

(a) 30°

(b) 45°

(c) 60°

(d) 90°

**Solution:**

**(d)** We have, AT = TP and TB = TP (Lengths of the tangents from ext. point T to the circles)

∠TAP = ∠TPA = x (say)

and ∠TBP = ∠TPB = y (say)

Also, in triangle APB,

x + x + x + y + y = 180°

=> 2x + 2y = 180°

=> x + y = 90°

=> ∠APB = 90°

**Question 41.**

In the figure, PQ and PR are two tangents to a circle with centre O. If ∠QPR= 46, then ∠QOR equals

(a) 67°

(b) 134°

(c) 44°

(d) 46°

**Solution:**

**(b)** ∠OQP = 90°

[Tangent is ⊥ to the radius through the point of contact]

∠ORP = 90°

∠OQP + ∠QPR + ∠ORP + ∠QOR = 360° [Angle sum property of a quad.]

90° + 46° + 90° + ∠QOR = 360°

∠QOR = 360° – 90° – 46° – 90° = 134°

**Question 42.**

In the 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 **[CBSE2014]**

(a) 3.8

(b) 7.6

(c) 5.7

(d) 1.9

**Solution:**

**(b)** In the figure, QR is common tangent to the two circles touching each other externally at T

Tangent at T meets QR at P

PT = 3.8 cm

PT and PQ are tangents from P

PT = PQ = 3.8 cm

Similarly PT and PR are tangents

PT = PR = 3.8 cm

QR = 3.8 + 3.8 = 7.6 cm

**Question 43.**

In the 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)**

**Solution:**

**(b)** In the given figure,

ABCD is a quadrilateral circumscribe a circle and its sides AB, BC, CD and DA touch the circle at P, Q, R and S respectively

AB = x cm, BC = 7 cm, CR = 3 cm, AS = 5 cm

CR and CQ are tangents to the circle from C

CR = CQ = 3 cm

BQ = BC – CQ = 7 – 3 = 4 cm

BQ = and BP are tangents from B

BP = BQ = 4 cm

AS and AP are tangents from A

AP = AS = 5 cm

AB = AP + BP = 5 + 4 = 9 cm

x = 9 cm

**Question 44.**

If angle between two radii of a circle is 130°, the angle between the tangent at the ends of radii is **(NCERT Exemplar)**

(a) 90°

(b) 50°

(c) 70°

(d) 40°

**Solution:**

**(b)** O is the centre of the circle.

Given, ∠POQ = 130°

PT and QT are tangents drawn from external point T to the circle.

∠OPT = ∠OQT = 90° [Radius is perpendicular to the tangent at point of contact]

In quadrilateral OPTQ,

∠PTQ + ∠OPT + ∠OQT + ∠POQ = 360°

=> ∠PTQ + 90° + 90° + 130° = 360°

=> ∠PTQ = 360° – 310° = 50°

Thus, the angle between the tangents is 50°.

**Question 45.**

If two tangents inclined at a angle of 60° are drawn to a circle of radius 3 cm, then length of each tangent is equal to **[NCERT Exemplar]**

(a) cm

(b) 6 cm

(c) 3 cm

(d) 3√3 cm

**Solution:**

**(d)** Let P be an external point and a pair of tangents is drawn from point P and angle between these two tangents is 60°.

Join OA and OP.

Also, OP is a bisector of line ∠APC

∠APO = ∠CPO = 30°

Also, OA ⊥ AP

Tangent at any point of a circle is perpendicular to the radius through the point of contact.

Hence, the length of each tangent is 3√3 cm

**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 **[NCERT Exemplar]**

(a) 3 cm

(b) 6 cm

(c) 9 cm

(d) 1 cm

**Solution:**

**(b)** Let O be the centre of two concentric circles C_{1} and C_{2}, whose radii are r_{1} = 4 cm and r_{2} = 5 cm.

Now, we draw a chord AC of circle C_{2}, which touches the circle C_{1} at B.

Also, join OB, which is perpendicular to AC. [Tangent at any point of circle is perpendicular to radius throughly the point of contact]

Now, in right angled ∆OBC, by using Pythagoras theorem,

OC² = BC² + BO² [(hypotenuse)² = (base)² + (perpendicular)²]

=> 5² = BC² + 4²

=> BC² = 25 – 16 = 9

=> BC = 3 cm

Length of chord AC = 2 BC = 2 x 3 = 6 cm

**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 **[NCERT Exemplar]**

(a) 4 cm

(b) 5 cm

(b) 6 cm

(d) 8 cm

**Solution:**

**(d)** First, draw a circle of radius 5 cm having centre O.

A tangent XY is drawn at point A.

A chord CD is drawn which is parallel to XY and at a distance of 8 cm from A.

Now, ∠OAY = 90°

[Tangent and any point of a circle is perpendicular to the radius through the point of contact]

∠OAY + ∠OED = 180°

[sum of cointerior is 180°]

=> ∆OED = 180°

Also, AE = 8 cm, Join OC

Now, in right angled ∆OBC

OC² = OE² + EC²

=> EC² = OC² – OE² [by Pythagoras theorem]

EC² = 5² – 3² [OC = radius = 5 cm, OE = AE – AO = 8 – 5 = 3 cm]

EC² = 25 – 9 = 16

=> EC = 4 cm

Hence, length of chord CD = 2 CE = 2 x 4 = 8 cm

[Since, perpendicular from centre to the chord bisects the chord]

**Question 48.**

From a point P which is at a distance 13 cm from the centre O of a circle of radius 5 cm, the pair of tangent PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is **[NCERT Exemplar]**

(a) 60 cm²

(b) 65 cm²

(c) 30 cm²

(d) 32.5 cm²

**Solution:**

**(a)** Firstly, draw a circle of radius 5 cm having centre O.

P is a point at a distance of 13 cm from O.

A pair of tangents PQ and PR are drawn.

Thus, quadrilateral PQOR is formed.

OQ ⊥ QP [since, AP is a tangent line]

In right angled ∆PQO,

OP² = OQ² + QP²

=> 13² = 5² + QP²

=> QP² = 169 – 25 = 144 = 12²

=> QP = 12 cm

Now, area of ∆OQP = x QP x QO = x 12 x 5 = 30 cm²

Area of quadrilateral QORP = 2 ∆OQP = 2 x 30 = 60 cm²

**Question 49.**

If PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to

(a) 25°

(b) 30°

(c) 40°

(d) 50°

**Solution:**

**(a)** Given, PA and PB are tangent lines.

PA = PB [Since, the length of tangents drawn from an ∠PBA = ∠PAB = θ [say]

In ∆PAB,

∠P + ∠A + ∠B = 180°

[since, sum of angles of a triangle = 180°

50°+ θ + θ = 180°

2θ = 180° – 50° = 130°

θ = 65°

Also, OA ⊥ PA

[Since, tangent at any point of a circle is perpendicular to the radius through the point of contact]

∠PAO = 90°

=> ∠PAB + ∠BAO = 90°

=> 65° + ∠BAO = 90°

=> ∠BAO = 90° – 65° = 25°

**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 **[NCERT Exemplar]**

(a) 10 cm

(b) 7.5 cm

(c) 5 cm

(d) 2.5 cm

**Solution:**

**(c)**

**Question 51.**

In the figure, if ∠AOB = 125°, then ∠COD is equal to **[NCERT Exemplar]**

(a) 45°

(b) 35°

(c) 55°

(d) 62°

**Solution:**

**(c)** We know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

∠AOB + ∠COD = 180°

=> ∠COD = 180° – ∠AOB = 180° – 125° = 55°

**Question 52.**

In the figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to **[NCERT Exemplar]**

(a) 20°

(b) 40°

(c) 35°

(d) 45°

**Solution:**

**(b)** Given, AB || PR

∠ABQ = ∠BQR = 70° [alternate angles]

Also QD is perpendicular to AB and QD bisects AB.

In ∆QDA and ∆QDB

∠QDA = ∠QDB [each 90°]

AD = BD

QD = QD [common side]

∆ADQ = ∆BDQ [by SAS similarity criterion]

Then, ∠QAD = ∠QBD …(i) [c.p,c.t.]

Also, ∠ABQ = ∠BQR [alternate interior angle]

∠ABQ = 70° [∠BQR = 70°]

Hence, ∠QAB = 70° [from Eq. (i)]

Now, in ∆ABQ,

∠A + ∠B + ∠Q = 180°

=> ∠Q = 180° – (70° + 70°) = 40°

Hope given RD Sharma Class 10 Solutions Chapter 8 Circles MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

## Leave a Reply