## RS Aggarwal Class 10 Solutions Chapter 12 Circles MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 12 Circles MCQS.

**Other Exercises**

- RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12A
- RS Aggarwal Solutions Class 10 Chapter 12 Circles Ex 12B
- RS Aggarwal Solutions Class 10 Chapter 12 Circles MCQS
- RS Aggarwal Solutions Class 10 Chapter 12 Circles Test Yourself

**Choose the correct answer in each of the following questions.**

**Question 1.**

**The number of tangents that can be drawn from an external point to a circle is [CBSE 2011, 12]**

**(a) 1**

**(b) 2**

**(c) 3**

**(d) 4**

**Solution:**

Number of tangents drawn from an external point to a circle is 2. (b)

**Question 2.**

**In the given figure, RQ is a tangent to the circle with centre O. If SQ = 6 cm and QR = 4 cm, then OR is equal to [CBSE 2014]**

**(a) 2.5 cm**

**(b) 3 cm**

**(c) 5 cm**

**(d) 8 cm**

**Solution:**

In the given figure, RQ is tangent to the circle with centre O.

SQ = 6 cm, QR = 4 cm

OR = √(OQ² + QR²) (In right ∆OQR)

**Question 3.**

**In a circle of radius 7 cm, tangent PT is drawn from a point P such that PT = 24 cm. If O is the centre of the circle, then length OP = ?**

**(a) 30 cm**

**(b) 28 cm**

**(c) 25 cm**

**(d) 18 cm**

**Solution:**

In the given figure, PT is tangent to the circle with centre O and radius OT = 7 cm, PT = 24 cm

OT is the radius and PT is the tangent OT ⊥ PT

Now, in right ∆OTP,

OP² = OT² + PT²

OP² = (7)² + (24)²

OP² = 49 + 576 = 625 = (25)²

OP = 25 cm (c)

**Question 4.**

**Which of the following pairs of lines in a circle cannot be parallel?**

**(a) two chords**

**(b) a chord and a tangent**

**(c) two tangents**

**(d) two diameters [CBSE2011]**

**Solution:**

Two diameters cannot be parallel. (d)

**Question 5.**

**The chord of a circle of radius 10 cm subtends a right angle at its centre. The length of the chord (in cm) is [CBSE 2014]**

**(a) **

**(b) 5√2**

**(c) 10√2**

**(d) 10√3**

**Solution:**

A chord subtends a right angle at its centre

Radius of the circle = 10 cm

**Question 6.**

**In the given figure, PT is a tangent to the circle with centre O. If OT = 6 cm and OP = 10 cm, then the length of tangent PT is**

**(a) 8 cm**

**(b) 10 cm**

**(c) 12 cm**

**(d) 16 cm**

**Solution:**

In the given figure, PT is tangent to the circle with centre O and radius

OT = 6 cm OP = 10 cm

OT is the radius and PT is the tangent

OT ⊥ TB

Now, in right ∆OPT,

OP² = OT² + PT² (Pythagoras Theorem)

⇒ (10)² = (6)² + PT²

⇒ 100 = 36 + PT²

⇒ PT² = 100 – 36 = 64 = (8)².

PT = 8 cm (a)

**Question 7.**

**In the given figure, 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 24 cm. Then, the radius of the circle is [CBSE 2011, 12]**

**(a)10 cm**

**(b) 12 cm**

**(c) 13 cm**

**(d) 15 cm**

**Solution:**

In the given figure, point P is 26 cm away from the centre O of the circle.

Length of tangent PT = 24 cm

Let radius = r

In right ∆OPT,

OP² = PT² + OT²

⇒ 26² = 24² + r²

⇒ r² = 26² – 24² = 676 – 576 = 100 = (10)²

r = 10

Radius = 10 cm (a)

**Question 8.**

**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 [CBSE 2014]**

**(a) 30°**

**(b) 45°**

**(c) 60°**

**(d) 90°**

**Solution:**

PQ is tangent to the circle with centre O at P. ∆OPQ is an isosceles triangle.

∠OQP = ?

∆OPQ is an isosceles triangle

OP = PQ

∠POQ = ∠OQP

But OP is radius and PQ is tangent

OP ⊥ PQ ⇒ ∠OPQ = 90°

∠POQ + ∠OQP = 90°

⇒ ∠POQ = ∠OQP = = 45°

Hence, ∠OQP = 45° (b)

**Question 9.**

**In the given figure, AB and AC are tangents to the circle with centre O such that ∠BAC = 40°. Then ∠BOC is equal to [CBSE 2011, 14]**

**(a) 80°**

**(b) 100°**

**(c) 120°**

**(d) 140°**

**Solution:**

In the given figure, AB and AC are tangents to the circle with centre O such that ∠BAC = 40°, ∠BOC = ?

AB and AC are tangents and OB and OC are radii.

OB ⊥ AB and OC ⊥ AC

⇒ ∠OBA = 90° and ∠OCA = 90°

In quadrilateral ∆BOC,

∠BAC + ∠BOC = 180°

⇒ 40° + ∠BOC = 180°

⇒ ∠BOC = 180° – 40° = 140° (d)

**Question 10.**

**If a chord AB subtends an angle of 60° at the centre of a circle, then the angle between the tangents to the circle drawn from A and B is [CBSE 2013C]**

**(a) 30°**

**(b) 60°**

**(c) 90°**

**(d) 120°**

**Solution:**

A chord AB subtends an angle of 60° at the centre of a circle with centre O.

TA and TB are tangents drawn to the circle.

Then, ∠ATB = 180° – ∠AOB = 180° – 60° = 120° (d)

**Question 11.**

**In the given figure, O is the centre of two concentric circles of radii 6 cm and 10 cm. AB is a chord of outer circle which touches the inner circle. The length of chord AB is**

**(a) 8 cm**

**(b) 14 cm**

**(c) 16 cm**

**(d) √136 cm**

**Solution:**

In the given figure, O is the centre of the two concentric circles of radii 6 cm and 10 cm.

AB is a chord of the outer circle and touches the inner circle at P.

OP = 6 cm, OA = 10 cm

OP is radius and APB is tangent to the inner circle.

OP ⊥ AB and P is the mid point of AB.

In right ∆OPA,

OA² = OP² + AP²

⇒ 10² = 6² + AP²

⇒ 100 = 36 + AP²

⇒ AP²= 100 – 36 = 64 = (8)²

AP = 8 cm

and AB = 2 x AP = 2 x 8 = 16 cm (c)

**Question 12.**

**In the given figure, AB and AC are tangents to a circle with centre O and radius 8 cm. If OA = 17 cm, then the length of AC (in cm) is [CBSE2012]**

**(a) 9**

**(b) 15**

**(c) √353**

**(d) 25**

**Solution:**

In the given figure, AB and AC are tangents to a circle with centre O and radius 8 cm.

OA = 17 cm.

**Question 13.**

**In the given figure, O is the centre of a circle, AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT = ?**

**(a) 40°**

**(b) 50°**

**(c) 60°**

**(d) 65°**

**Solution:**

In the given figure, O is the centre of the circle, AT is tangent, AOC is the diameter and ∠ACB = 50°

We have to find the measure of ∠BAT

AB is chord and AT is the tangent

∠ACB = ∠BAT (Angles in the alternate segment)

= 50° (b)

**Question 14.**

**In the given figure, O is the centre of a circle, PQ is a chord and PT is the tangent at P. If ∠POQ = 70°, then ∠TPQ is equal to [CBSE2011]**

**(a) 35°**

**(b) 45°**

**(c) 55°**

**(d) 70°**

**Solution:**

O is the centre of circle, PQ is a chord, PT is tangent.

∠POQ = 70°, then ∠TPQ = ?

Take a point R on the major segment and join PR and QR

arc PQ subtends ∠POQ at the centre and ∠PRQ at the remaining part of the circle

∠PRQ = ∠POQ = x 70° = 35°

But ∠TPQ = ∠PRQ (Angles in the alternate segment)

∠TPQ = 35° (a)

**Question 15.**

**In the given figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Then, AT = ?**

**(a) 4 cm**

**(b) 2 cm**

**(c) 2√3 cm**

**(d) 4√3 cm**

**Solution:**

In the given figure, AT is the tangent to the circle with centre O and OA is its radius OT = 4 cm, ∠OTA = 30°

Now, we have to find the length of AT

**Question 16.**

**If PA and PB are two tangents to a circle with centre O such that ∠AOB = 110°, then ∠APB = ? [CBSE 2011, 14]**

**(a) 55°**

**(b) 60°**

**(c) 70°**

**(d) 90°**

**Solution:**

In the given figure, PA and PB are the two tangents to the circle with centre O, which subtends ∠AOB = 110°

Now, we have to find the measure ∠APB

OA and OB are the radii of the circle and AP and BP are the tangents

OA ⊥ AP and OB ⊥ BP

∠A = ∠B = 90°

In quadrilateral OAPB,

∠A + ∠B = 90° + 90°= 180°

∠AOB + ∠APB = 180°

⇒ 110° + ∠APB = 180°

⇒ ∠APB = 180° – 110° = 70° (c)

**Question 17.**

**In the given figure, the length of BC is [CBSE 2012, 14]**

**(a) 7 cm**

**(b) 10 cm**

**(c) 14 cm**

**(d) 15 cm**

**Solution:**

In the given figure, in ∆ABC,

BC = ?

AF and AE are the tangents to the circle from A.

AE = AF = 4 cm CE = AC – AE = 11 – 4 = 7 cm

Similarly, CD and CE are tangents

CD = CE = 7 cm

and BF and BD are tangents BD = BF = 3 cm

BC = BD + CD = 3 + 7 = 10 cm (b)

**Question 18.**

**In the given figure, If ∠AOD = 135° then ∠BOC is equal to [CBSE 2013C]**

**(a) 25°**

**(b) 45°**

**(c) 52.5°**

**(d) 62.5°**

**Solution:**

In the given figure, ∠AOD = 135°

We know that if a circle is inscribed in a quadrilateral, the opposite sides subtends supplementary angles.

∠AOD + ∠BOC = 180°

135° + ∠BOC = 180°

⇒ ∠BOC = 180° – 135° = 45° (b)

**Question 19.**

**In the given figure, O is the centre of a circle and PT is the tangent to the circle. If PQ is a chord such that ∠QPT = 50°, then ∠POQ = ?**

**(a) 100°**

**(b) 90°**

**(c) 80°**

**(d) 75°**

**Solution:**

In the given figure, PQ is a chord of a circle with centre O and PT is a tangent at P to the circle such that ∠QPT = 50°.

Then, we have to find ∠POQ.

PT is the tangent and OP is the radius

OP ⊥ PT ⇒ ∠OPT = 90°

∠OPQ = ∠OPT – ∠QPT = 90° – 50° = 40°

In ∆OPQ,

OP = OQ (radii of the same circle)

∠OPQ = ∠OQP = 40°

and ∠POQ = 180° – (∠OPQ + ∠OQP)

= 180° – (40° + 40°)

= 180°- 80° = 100° (a)

**Question 20.**

**In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 60° then ∠OAB is [CBSE 2011]**

**(a) 15°**

**(b) 30°**

**(c) 60°**

**(d) 90°**

**Solution:**

In the given figure, PA and PB are two tangents to the circle with centre O.

∠APB = 60° then ∠OAB

Join OB.

PAOB is a cyclic quadrilateral.

∠APB + ∠AOB = 180°

OA is radius and PA is tangent

OA ⊥ AP ⇒ ∠OAP = 90°

PA = PB (Tangents to the circle)

∠PAB = ∠PBA

But, ∠PAB + ∠PBA = 180° – 60° = 120°

∠PAB = ∠PBA = = 60°

∠OAB = 90° – 60° = 30° (b)

**Question 21.**

**If two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm, then the length of each tangent is**

**(a) 3 cm**

**(b) cm**

**(c) 3√3 cm**

**(d) 6 cm**

**Solution:**

Two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm.

Join OA, OB and OP.

**Question 22.**

**In the given figure, PQ and PR are tangents to a circle with centre A. If ∠QPA = 21° then ∠QAR equals [CBSE2012]**

**(a) 63°**

**(b) 117°**

**(c) 126°**

**(d) 153°**

**Solution:**

In the given figure, PQ and PR are tangents drawn from an external point P to a circle with centre A.

∠QPA = 27°, ∠QAR = ?

AP bisects ∠QPR and ∠QPA = 27°

∠QPR = 2 x 27° = 54°

But ∠QPR + ∠QAR = 180° (QARP is a cyclic quadrilateral)

⇒ 54° + ∠QAR = 180°

⇒ ∠QAR = 180° – 54° = 126° (c)

**Question 23.**

**In the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm. If PA ⊥ PB, then the length of each tangent is [CBSE 2013]**

**(a) 3 cm**

**(b) 4 cm**

**(c) 5 cm**

**(d) 6 cm**

**Solution:**

In the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm.

PA ⊥ PB, then length of tangent is = ?

Join GA and CB.

CAand CB are radii and PA, PB are tangents to the circle.

CA ⊥ PA and CB ⊥ PB But, ∠APB = 90°

∠ACB = 180° – 90° = 90°

PA = PB tangents of a circle

CAPB is a square

PA = PB = radius = 4 cm (b)

**Question 24.**

**If PA and PB are two tangents to a circle with centre O such that ∠APB = 80°. Then, ∠AOP = ?**

**(a) 40°**

**(b) 50°**

**(c) 60°**

**(d) 70°**

**Solution:**

In the given figure, PA and PB are tangents to the circle with centre O such that ∠APB = 80°.

Join OP

Now, in right ∆OAP, ∠A = 90°

∠AOP = 90° – 40° = 50° (b)

**Question 25.**

**In the given figure, O is the centre of the circle. AB is the tangent to the circle at the point P. If ∠APQ = 58° then the measure of ∠PQB is [CBSE2014]**

**(a) 32°**

**(b) 58°**

**(c) 122°**

**(d) 132°**

**Solution:**

In the given figure, O is the centre of a circle. AB is the tangent to the circle at point P.

∠APQ = 58°, ∠PQB = ?

∠QPR = 90° (Angle in a semi circle)

But, ∠RPB + ∠QPR + ∠APQ = 180° (Angles on one side of a line)

⇒ ∠RPB + 90° + 58° = 180°

⇒ ∠RPB + 148° = 180°

⇒ ∠RPB = 180° – 148° = 32°

∠PQR or ∠PQB = ∠RPB (Angles in the alternate segment)

⇒ ∠PQB = 32° (a)

**Question 26.**

**In the given figure, O is the centre of the circle. AB is the tangent to the circle at the point P. If ∠PAO = 30° then ∠CPB + ∠ACP is equal to**

**(a) 60°**

**(b) 90°**

**(c) 120°**

**(d) 150°**

**Solution:**

In the given figure, O is the centre of the circle. AB is tangent to the circle at P.

∠PAO = 30°

∠CPB + ∠ACP = ?

∠CPD = 90° (Angle in a semi circle)

∠DPA + ∠CPB = 90°

But, ∠DPA = ∠ACP (Angles in alternate segment)

∠CPB + ∠ACP = 90° (b)

**Question 27.**

**In the given figure, PQ is a tangent to a circle with centre O. A is the point of contact. If ∠PAB = 67°, then the measure of ∠AQB is**

**(a) 73°**

**(b) 64°**

**(c) 53°**

**(d) 44°**

**Solution:**

In the given figure, PQ is the tangent to the circle at A.

∠PAB = 67°, ∠AQB = ?

Join BC.

∠BAC = 90° (Angle in a semi circle)

But, ∠PAB + ∠BAC + ∠CAQ = 180°

⇒ 67° + 90° + ∠CAQ = 180°

⇒ 157° + ∠CAQ = 180°

∠CAQ = 182° – 157° = 23°

∠ACB = ∠PAB (Angles in the alternate segment)

∠ACB = 67°

In ∆ACQ,

Ext. ∠ACB = ∠CAQ + ∠AQC

⇒ 67° = 23° + ∠AQC

⇒ ∠AQC = 67° – 23° = 44°

⇒ ∠AQB = 44° (d)

**Question 28.**

**In the given figure, two circles touch each other at C and AB is a tangent to both the circles. The measure of ∠ACB is [CBSE 2013C] [HOTS]**

**(a) 45°**

**(b) 60°**

**(c) 90°**

**(d) 120°**

**Solution:**

In the given figure, two circles touch each other at C. AB is the common tangent.

∠ACB = ?

Draw a tangent from C which meets AB at P.

PA and PC are tangents to the first circle.

PA = PC

∠PAC = ∠PCA …(i)

Similarly, PB = PC

∠PCB = ∠PBC …(ii)

Adding, ∠PAC + ∠PBC = ∠PCA + ∠PCB

⇒ ∠PAC + ∠PBC = ∠ACB

But, ∠PAC + ∠PBC + ∠ACB = 180° (Angles of a triangle)

∠ACB = 90° (c)

**Question 29.**

**O is the centre of a circle of radius 5 cm. At a distance of 13 cm from O, a point P is taken. From this point, two tangents PQ and PR are drawn to the circle. Then, the area of quad. PQOR is**

**(a) 60 cm²**

**(b) 32.5 cm²**

**(c) 65 cm²**

**(d) 30 cm²**

**Solution:**

In the given figure O is the centre of the circle with radius 5 cm P is a point out side the circle and OP = 13 cm

PQ and PR are the tangents to the circle drawn from P

We have to find the area of quad. PQOR

OQ is radius and PQ is the tangent

OQ ⊥ QP

In ∆OPQ,

OP² = OQ² + PQ² (Pythagoras Theorem)

⇒ (13)² = (5)² + PQ²

⇒ 169 = 25 + PQ²

⇒ PQ² = 169 – 25 = 144 = (12)²

PQ = 12 cm

PQ = PR = 12 cm

Now, diagonal OP bisects the quad. PQOR into two triangles equal in areas.

Now, area of ∆PQO = x PQ x OQ

= x 12 x 5 = 30 cm²

Area of quad. PQOR = 2 x area ∆PQO = 2 x 30 = 60 cm² (a)

**Question 30.**

**In the given figure, PQR is a tangent to the circle at Q, whose centre is O and AB is a chord parallel to PR such that ∠BQR = ****70°. Then, ∠AQB = ?**

**(a) 20°**

**(b) 35°**

**(c) 40°**

**(d) 45°**

**Solution:**

In the given figure,

PQR is a tangent drawn at Q to the circle with centre O.

AB is a chord parallel to PR such that ∠BQR = 70°

Then, we have to find ∠AQB

Join QO and produce it to AB meeting it at L.

OQ ⊥ PR ⇒ LQ ⊥ PR

QL bisects AB at L

QA = QB

∆QAB is an isosceles triangle

∠LQA = ∠LQB

∠LQA = ∠LQR – ∠BQR = 90° – 70° = 20°

∠AQB = 2 x 20° = 40° (c)

**Question 31.**

**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 [CBSE 2013C]**

**(a) 8 cm**

**(b) √104 cm**

**(c) 12 cm**

**(d) √125 cm**

**Solution:**

Length of a tangent to the circle from an external point = 10 cm

Radius (r) = 5 cm OP = ?

OQ is radius and QP is tangent

OQ ⊥ QP

In right ∆OPQ,

OP² = OQ² + QP² (Pythagoras Theorem) = (5)² + (10)² = 25 + 100 = 125

OP = √125 cm (d)

**Question 32.**

**In the given figure, O is the centre of a circle, BOA is its diameter and the tangent at the point P meets BA extended at T. If ∠PBO = 30°, then ∠PTA = ?**

**(a) 60°**

**(b) 30°**

**(c) 15°**

**(d) 45°**

**Solution:**

In the figure, O is the centre of the circle BOA is its diameter and PT is tangent at P which meets BA produced at T. ∠PBO = 30°.

We have to find ∠PTA

In ∆BOP,

OB = OP (radii of the same circle)

∠APB = ∠PBO = 30°

But, OP is radius and PT is the tangent

OP ⊥ PT ⇒ ∠OPT = 90°

∠BPT = ∠BPO + ∠OPT = 30° + 90° = 120°

Now, in ∆PBT,

∠BPT + ∠PBA + ∠PTA = 180° (sum of angles of a triangle)

⇒ 120° + 30° + ∠PTA = 180°

⇒ 150° + ∠PTA = 180°

⇒ ∠PTA = 180° – 150° = 30° (b)

**Question 33.**

**In the given figure, a circle touches the side DF of ∆EDF 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) 9 cm**

**(b) 12 cm**

**(c) 13.5 cm**

**(d) 18 cm**

**Solution:**

In the given figure, a circle touches the side DF of a AEDF at H and touches ED and EF on producing at K and M respectively.

EK = 9 cm.

Perimeter of ∆EDF.

DH and DK are tangents to the circle.

DH = DK

Similarly, ∠FH = ∠FM and EK = EH = 9 cm

EK = ED + DK ⇒ ED + DH = 9 cm…(i)

Similarly, EH = EF = FH = EF + FM = 9 cm …(ii)

Adding (i) and (ii)

ED + DH + EF + FH = 9 + 9 cm (DH + HF = DF)

ED + DF + FE = 18 cm

Perimeter of ∆EDF = 18 cm (d)

**Question 34.**

**To draw a pair of tangents to a circle, which are inclined to each other at an angle of 45°, we have to draw tangents at the end points of those two radii, the angle between which is [CBSE2011]**

**(a) 105°**

**(b) 135°**

**(c) 140°**

**(d) 145°**

**Solution:**

In the given figure, PA and PB are two tangents drawn from an external point P which inclined at an angle of 45°.

OA and OB are radii of the circle.

To find ∠AOB

AOBP is a cyclic quadrilateral

∠AOB + ∠APB = 180°

⇒ ∠AOB + 45° = 180°

⇒ ∠AOB = 180° – 45° = 135° (b)

**Question 35.**

**In the given figure, O is the centre of a circle; PQL and PRM are the tangents at the points Q and R respectively and S is a point on the circle such that ∠SQL = 50° and ∠SRM = 60°. Then, ∠QSR = ?**

**(a) 40°**

**(b) 50°**

**(c) 60°**

**(d) 70°**

**Solution:**

In the given figure, O is the centre of the circle PQL and PRM are the tangents from P drawn to the circle meeting it at Q and R respectively ∠SQL = 50°, and ∠SRM = 60°

Now, we have to find ∠QSR,

Join OQ, OR and OS OQ is radius and QP is tangent

OQ ⊥ QP

Similarly, OR ⊥ RP

∠1 = 90° – 50° = 40° and ∠2 = 90° – 60° = 30°

OS = OQ (radii of the same circle)

∠3 = ∠1 = 40°

Similarly OS = OR

∠2 = ∠4 = 30°

∠QSR = ∠3 + ∠4 = 40° + 30° = 70° (d)

**Question 36.**

**In the given figure, a triangle PQR is drawn to circumscribe a circle of radius 6 cm such that the segments QT and TR into which QR is divided by the point of contact T, are of lengths 12 cm and 9 cm respectively. If the area of ∆PQR = 189 cm² then the length of side PQ is [CBSE 2011]**

**(a) 17.5 cm**

**(b) 20 cm**

**(c) 22.5 cm**

**(d) 25 cm**

**Solution:**

In the given figure, a ∆PQR is drawn to inscribe a circle with centre O and radius 6 cm.

OT is radius.

QT = 12 cm, TR = 9 cm

Area ∆PQR =189 cm²

PQ = ?

QK = QT = 12 cm

RS = RT = 9 cm

Let PK = PS = x cm

PQ = 12 + x

PR = 9 + x cm

Area of A = x r x Perimeter of ∆PQR

⇒ 189 = x 6 x (PQ + QR + RP)

⇒ 189 = 3 (12 + x + 21 + 9 + x)

⇒ 63 = 42 + 2x

⇒ 2x = 63 – 42 = 21

x = 10.5

AB = 10.5 + 12 = 22.5 cm (c)

**Question 37.**

**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 is [CBSE 2014]**

**(a) 1.9 cm**

**(b) 3.8 cm**

**(c) 5.7 cm**

**(d) 7.6 cm**

**Solution:**

In the given figure, QR is a common tangent to two given circles touching each other externally at point T.

A tangent PT is drawn from T which intersects QR at P.

PT = 3.8 cm, QR = ?

PT and PQ are tangents to the first circle.

PQ = PT …(i)

Similarly, PT and PR tangents to the second circle.

PR = PT …(ii)

From (i) and (ii),

PQ = PR = PT = 3.8 cm

QR = 3.8 + 3.8 = 7.6 cm (d)

**Question 38.**

**In the given figure, quad. ABCD is circumscribed, touching the circle at P, Q, R and S. If AP = 5 cm, BC = 7 cm and CS = 3 cm. Then, the length AB = ?**

**(a) 9 cm**

**(b) 10 cm**

**(c) 12 cm**

**(d) 8 cm**

**Solution:**

In the figure, quadrilateral ABCD is circumscribed touches the circle at P, Q, R and S

AP = 5 cm, BC = 7 cm, CS = 3 cm AB = ?

Tangents drawn from the external point to the circle are equal

AQ = AP = 5 cm

CR = CS = 3 cm

BQ = BR

Now, BR = BC – CR = 7 – 3 = 4 cm

BQ = 4 cm

Now, AB = AQ + BQ = 5 + 4 = 9 cm (a)

**Question 39.**

**In the given figure, quad. ABCD is circumscribed, touching the circle at P, Q, R and S. If AP = 6 cm, BP = 5 cm, CQ = 3 cm and DR = 4 cm, then perimeter of quad. ABCD is**

**(a) 18 cm**

**(b) 27 cm**

**(c) 36 cm**

**(d) 32 cm**

**Solution:**

In the given figure, quad. ABCD is circumscribed touching the circle at P, Q, R and S

AP = 6 cm, BP = 5 cm, CQ = 3 cm and DR = 4 cm.

Now, we have to find the perimeter of the quad. ABCD.

We know that tangents from an external point to the circle are equal.

AP = AS = 6 cm

BP = BQ = 5 cm

CQ = CR = 3 cm

DR = DS = 4 cm

AB = AP + BP = 6 + 5 = 11 cm

BC = BQ + CQ = 5 + 3 = 8 cm

CD = CR + DR = 3 + 4 = 7 cm

and DA = AS + DS = 6 + 4 = 10 cm

Perimeter of the quad. ABCD

= AB + BC + CD + DA

= (11 + 8 + 7 + 10) cm

= 36 cm (c)

**Question 40.**

**In the given figure, O is the centre of a circle, AB is a chord and AT is the tangent at A. If ∠AOB = 100° then ∠B AT is equal to [CBSE2011]**

**(a) 40°**

**(b) 50°**

**(c) 90°**

**(d) 100°**

**Solution:**

In the given figure, O is the centre of the circle, AB is chord and AT is the tangent at A.

∠AOB = 100°, ∠BAT = ?

Take a point P on the major segment of the circle and join AP and BP.

Arc AB subtends ∠AOB at the centre and ∠APB at the remaining part of the circle.

∠APB = ∠AOB = x 100° = 50°

Now, ∠BAT = ∠APB (Angles in the alternate segment)

∠BAT = 50° (b)

**Question 41.**

**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 is [CBSE 2014]**

**(a) 1 cm**

**(b) 2 cm**

**(c) 3 cm**

**(d) 4 cm**

**Solution:**

In a right ∆ABC, right angled at B

BC = 12 cm, AB = 5 cm

A circle is inscribed in it touching its sides at P, Q and R.

Join OP, OQ and OR.

AC² = AB² + BC² (Pythagoras Theorem)

= 5² + 12² = 25 + 144 = 169 = (13)²

**Question 42.**

**In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides AB, BC, CD and AD at P, Q, R and S respectively. If the radius of the circle is 10 cm, BC = 38 cm, PB = 27 cm and AD ⊥ CD then the length of CD is [CBSE 2013]**

**(a) 11 cm**

**(b) 15 cm**

**(c) 20 cm**

**(d) 21 cm**

**Solution:**

In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides, AB, BC, CD and DA at P, Q, R and S respectively

Radius OS = 10 cm

BC = 38 cm, PB = 27 cm

AD ⊥ DC

Length of CD = ?

Join OR and OS

BP and BQ are tangents to the circle

BQ = BP = 27 cm

BC = 38 cm

QC = 38 – 27 = 11 cm

CQ and CR are the tangents to the circle

CR = CQ = 11 cm

DR and DS are the tangents to the circle

DR = DS

AD ⊥ CD

OS is the radius and AD is the tangent

OS ⊥ AD

Similarly, OR ⊥ DC

OSDR is a square whose each side is equal to the radius = 10 cm

DR = DS = 10 cm

CD = CR + DR = 11 + 10 = 21 cm (d)

**Question 43.**

**In the given figure, AABC is right-angled at B such that BC = 6 cm and AB = 8 cm. A circle with centre O has been inscribed inside the triangle. OP ⊥ AB, OQ ⊥ BC and OR ⊥ AC. If OP = OQ = OR = x cm, then x = ?**

**(a) 2 cm**

**(b) 2.5 cm**

**(c) 3 cm**

**(d) 3.5 cm**

**Solution:**

In the given figure, ∆ABC is a right angled triangle, right angle at ∠B.

BC = 6 cm, AB = 8 cm

A circle with centre O is inscribed inside the triangle ABC

OP ⊥ AB and OQ ⊥ BC and OR ⊥ AC

OP = OQ = OR = x cm

OPBQ is a square

In right ∆ABC,

AC² = AB² + BC² (Pythagoras Theorem)

= (8)² + (6)² = 64 + 36 = 100 = (10)²

AC = 10 cm

Tangents drawn from the external point to the circle are equal

BP = BQ = x

CQ = CR = 6 – x

AP = AR = 8 – x

AR + CR = AC

⇒ 8 – x + 6 – x = 10

⇒ 14 – 2x = 10

⇒ 2x = 14 – 10 = 4

x = 2

Hence r = 2cm (a)

**Question 44.**

**Quadrilateral ABCD is circumscribed to a circle. If AB = 6 cm, BC = 7 cm and CD = 4 cm then the length of AD is [CBSE 2012]**

**(a) 3 cm**

**(b) 4 cm**

**(c) 6 cm**

**(d) 7 cm**

**Solution:**

A quadrilateral ABCD is circumscribed to a circle with centre O.

AB = 6 cm, BC = 7 cm, CD = 4 cm, AD = 7 cm

ABCD circumscribed to a circle.

AB + CD = BC + AD

⇒ 6 + 4 = 7 + AD

⇒ 10 = 7 + AD

AD = 10 – 7 = 3 cm (a)

**Question 45.**

**In the given figure, PA and PB are tangents to the given circle such that PA = 5 cm and ∠APB = 60°. The length of chord AB is**

**(a) 5√2 cm**

**(b) 5 cm**

**(c) 5√3 cm**

**(d) 7.5 cm**

**Solution:**

In the given figure, PA and PB are the tangents to the circle with centre O from P

PA = 5 cm, ∠APB = 60°

PA = PB = 5 cm

In ∆APB, ∠P = 60° and PA = PB

PAB is an equilateral triangle

AB = AP = BP = 5 cm (b)

**Question 46.**

**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 [CBSE 2013]**

**(a) 3 cm**

**(b) 4 cm**

**(c) 5 cm**

**(d) 6 cm**

**Solution:**

In the given figure, DE and DF are tangents to the circle from an external point D.

A is the centre of the circle.

DF = 5 cm and DE ⊥ DF, radius of the circle = 3

Join EA and FA.

AE and AF are the radius of the circle and DE and BF are the tangents.

AE ⊥ DE and AF ⊥ DF

∠EAF = 180° – ∠EDF = 180° – 90° = 90°

AEDF is a square.

AE = 5 cm

Radius of the circle = 5 cm (c)

**Question 47.**

**In the given figure, three circles with centres A, B, C respectively touch each other externally. If AB = 5 cm, BC = 7 cm and CA = 6 cm, then the radius of the circle with centre A is**

**(a) 1.5 cm**

**(b) 2 cm**

**(c) 2.5 cm**

**(d) 3 cm**

**Solution:**

In the given figure, three circles with centre A, B and C are drawn touching each other externally

AB = 5 cm, BC = 7 cm and CA = 6 cm

Let r1, r2, r3 be the radii of three circles respectively

**Question 48.**

**In the given figure, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC = 4 cm then the length of AP is [CBSE2012]**

**(a) 15 cm**

**(b) 10 cm**

**(c) 9 cm**

**(d) 7.5 cm**

**Solution:**

In the given figure, AP, AQ and BC are tangents to the circle.

AB = 5 cm, AC = 6 cm, BC = 4 cm

Length of AP = ?

BP and BR are the tangents to the circle.

BP = BR

Similarly, CR and CQ are tangents

CR = CQ

S and AP and AQ are tangents

AP = AQ

AP = AB + BP = AB + BR

AQ = AC + CQ = AC + CR

AP + AQ = AB + BR + AC + CR = AB + BR + CR + AC

AP + AP = AB + BC + AC

2AP = 5 + 4 + 6 = 15 cm

AP = = 7.5 cm (d)

**Question 49.**

**In the given figure, O is the centre of two concentric circles of radii 5 cm and 3 cm. From an external point P tangents PA and PB are drawn to these circles. If PA = 12 cm, then PB is equal to**

**(a) 5√2 cm**

**(b) 3√5 cm**

**(c) 4√10 cm**

**(d) 5√10 cm**

**Solution:**

In the given figure, O is the centre of two concentric circles of radii 5 cm and 3 cm respectively.

From external point P, PA and PB are tangents are drawn to the external circle and internal circle respectively

PA = 12 cm, PB = ?

OA and OB are the radii

OA ⊥ AP and OB ⊥ BP

Now, in right ∆OAP,

OP² = OA² + AP² (Pythagoras Theorem)

= (5)² + (12)²

= 25 + 144 = 169 = (13)²

OP = 13 cm

and in right ∆OBP,

OP² = OB² + BP²

(13)² = (3)² + BP²

⇒ 169 = 9 + BP²

⇒ PB² = 169 – 9 = 160

PB = √160 = √(16 x 10) = 4√10 cm (c)

**True/False Type**

**Question 50.**

**Which of the following statements is not true ?**

**(a) If a point P lies inside a circle, no tangent can be drawn to the circle, passing through P.**

**(b) If a point P lies on the circle, then one and only one tangent can be drawn to the circle at P.**

**(c) If a point P lies outside the circle, then only two tangents can be drawn to the circle from P.**

**(d) A circle can have more than two parallel tangents, parallel to a given line.**

**Solution:**

(a) It is true that no tangent can be drawn from a point inside the circle.

(b) It is true, that one and only one tangent can be drawn from a point on the circle.

(c) True. If a point P is outside the circle, two tangents can be drawn to the circle.

(d) No, only two parallel tangents can be drawn which are parallel to a given line. (d)

**Question 51.**

**Which of the following statements is not true ?**

**(a) A tangent to a circle intersects the circle exactly at one point.**

**(b) The point common to the circle and its tangent is called the point of contact.**

**(c) The tangent at any point of a circle is perpendicular to the radius of the circle through the point of contact.**

**(d) A straight line can meet a circle at one point only.**

**Solution:**

(a) It is true as a tangent intersects (touches) the circle exactly at one point.

(b) It is true that common point to the circle where the tangent touches the circle is called point of contact.

(c) It is true that the radius through the point of contact of a tangent is perpendicular to it.

(d) False as a straight line can meet at the most two points. (d)

**Question 52.**

**Which of the following statements is not true ?**

**(a) A line which intersects a circle in two points, is called a secant of the circle.**

**(b) A line intersecting a circle at one point only, is called a tangent to the circle.**

**(c) The point at which a line touches the circle, is called the point of contact.**

**(d) A tangent to the circle can be drawn from a point inside the circle.**

**Solution:**

(a) It is true, that a secant is a line which intersects the circle at two points.

(b) It is true, as a tangent intersects the circle at only one point.

(c) It is true that the point at which a tangent touches the circle is called a point of contact.

(d) It is false, as no tangent can be drawn from a point in side the circle.

**Assertion and Reason Type**

**Each question consists of two statements, namely, Assertion (A) and Reason (R).**

**For selecting the correct answer, use the following code :**

**(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).**

**(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).**

**(c) Assertion (A) is true and Reason (R) is false.**

**(d) Assertion (A) is false and Reason (R) is true.**

**Question 53.**

**Assertion (A):**

**At a point P of a circle with centre O and radius 12 cm, a tangent PQ of length 16 cm is drawn. Then, OQ = 20 cm.**

**Reason (R):**

**The tangent at any point of a circle is perpendicular to the radius through the point of contact.**

**The correct answer is : (a) / (b) / (c) / (d).**

**Solution:**

**In Assertion (A):**

In right ∆OPQ, OP ⊥ PQ

OQ² = OP² + PQ² = (12)² + (16)² = 144 + 256 = 400 = (20)²

OQ = 20 cm, which is true

**In Reason (R):**

It is also with respect to (A) (a)

**Question 54.**

**Assertion (A):**

**If two tangents are drawn to a circle from an external point, then they subtend equal angles at the centre.**

**Reason (R):**

**A parallelogram circumscribing a circle is a rhombus.**

**The correct answer is : (a) / (b) / (c) / (d).**

**Solution:**

**Assertion (A):**

The statement is true

**In Reason (R):**

It is also true but not with respect to (A) (b)

**Question 55.**

**Assertion (A):**

**In the given figure, a quad. ABCD is drawn to circumscribe a given circle, as shown. Then, AB + BC = AD + DC.**

**Reason (R):**

**In two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.**

**The correct answer is : (a) / (b) / (c) / (d).**

**Solution:**

**In Assertion (A):**

In the figure, ABCD is a quad, which is circumscribed a given circle.

Sum of opposite sides are equal

AB + CD = BC + AD

It is not true that AB + BC = AD + DC

**In Reason (R):**

It is true but not with respect to Assertion (A) (d)

Hope given RS Aggarwal Solutions Class 10 Chapter 12 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.

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