We have provided you with Extra and Important Questions from Class 10 Maths Chapter 10 Circles. This Extra and Important Questions will help you to score 100% in your Board Exams. These extra questions will be helpful to revise the important topics and concepts.

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**Circles Class 10 Important Questions with Answers Maths Chapter 10**

## Extra Questions for Class 10 Maths Chapter 10 Very Short Answer Type

**Question: **If a point P is 17 cm from the centre of a circle of radius 8 cm, then find the length of the tangent drawn to the circle from point P.

**Solution: **

OA ⊥ PA (∵ radius is ⊥ to tangent at point of contact)

∴ In ∆OAP, we have

PO^{2} = PA^{2} + AO^{2}

⇒ (17)^{2} = (PA)^{2} + (8)^{2}

(PA)^{2} = 289 – 64 = 225

⇒ PA = √225 = 15

Hence, the length of the tangent from point P is 15 cm.

**Question: **The length of the tangent to a circle from a point P, which is 25 cm away from the centre, is 24 cm. What is the radius of the circle?

**Solution: **

∵ OQ ⊥ PQ

∴ PQ^{2} + QO^{2} = OP^{2}

⇒ 25^{2} = OQ^{2} + 24^{2}

or OQ = √625 – √576

= √49 = 7 cm

**Question: **In Fig. 8.6, ABCD is a cyclic quadrilateral. If ∠BAC = 50° and ∠DBC = 60° then find ∠BCD.

**Solution: **

Here ∠BDC = ∠BAC = 50° (angles in same segment are equal)

In ABCD, we have

∠BCD = 180° – (∠BDC + ∠DBC)

= 180° – (50° + 60°)= 70°

**Question: **In Fig. 8.7, the quadrilateral ABCD circumscribes a circle with centre O. If ∠AOB = 115°, then find ∠COD.

**Solution:**

∵ ∠AOB = ∠COD (vertically opposite angles)

∴ ∠COD = 115°

**Question: **

In Fig. 8.8, AABC is circumscribing a circle. Find the length of BC.

**Solution: **AN = AM = 3 cm [Tangents drawn from an external point]

BN = BL = 4 cm [Tangents drawn from an external point]

CL = CM = AC – AM = 9 – 3 = 6 cm

⇒ BC = BL + CL = 4 + 6 = 10 cm.

**Question: **In Fig. 8.9, O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ. Find ∠POQ.

**Solution: **

∠OPQ = 90° – 50° = 40°

OP = OQ [Radii of a circle]

∠OPQ = ∠OQP = 40°

(Equal opposite sides have equal opposite angles)

∠POQ = 180° – ∠OPQ – ∠OQP

= 180° – 40° – 40° = 100°

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

**Solution: **

OA = 4 cm, OB = 5 cm

Also, OA ⊥ BC

∴ OB^{2} = OA^{2} + AB^{2}

⇒ 5^{2} = 4^{2} + AB^{2}

⇒ AB = √25 – √16 = 3 cm

⇒ BC = 2 AB = 2 × 3 = 6 cm

**Question: **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 find ∠OPQ.

**Solution: **

∠OQP = 90°

∠QOP = 180° – 120° = 60°

∠OPQ = 180° – ∠OQP – ∠QOP

= 180° – 90° – 60°

= 30°

**Question: **

From an external point P, tangents PA and PB are drawn to a circle with centre O. If ∠PAB = 50°, then find ∠AOB.

**Solution: **n:

∵ PA = PB ⇒ ∠BAP = ∠ABP = 50°

∴ ∠APB = 180° – 50° – 50° = 80°

∴ ∠AOB = 180° – 80° = 100°

**Question: **In PQ is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°, find ∠PCA.

**Solution: **

∠ACB = 90° (Angle in the semicircle)

∠CAB = 30° (given)

In ∆ABC,

90° + 30° + ∠ABC = 180°

⇒ ∠ABC = 60°

Now, ∠PCA = ∠ABC (Angles in the alternate segment)

∴ ∠PCA = 60°

or

Construction: Join O to C.

∠PCO = 90° [∵ Line joining centre to point of contact is perpendicular to PQ]

In ∆AOC, OA = OC [Radii of circle]

∴ ∠OAC = ∠OCA = 30° [Equal sides have equal opp. angles]

Now, ∠PCA = ∠PCO – ∠ACO

= 90° – 30° = 60°

**Question: **In fig. 8.16, 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.

**Solution: **

PQ = PR = 5 cm [∵ Tangents drawn from external point are equal] PK

∴ PS = 2PQ = 10 cm [∵ Perpendicular drawn from centre to the chord bisects the chord]

## Extra Questions for Class 10 Maths Chapter 10 Short Answer Type

State true or false for each of the following and justify your answer (Q. 1 to 3)

**Question: **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.

**Solution: **

True, Join OC,

∠ACB = 90° (Angle in semi-circle)

∴ ∠OBC = 180o – (90° + 30°) = 60°

Since, OB = OC = radii of same circle [Fig. 8.16]

∴ ∠OBC = ∠OCB = 60°

Also, ∠OCD = 90°

⇒ ∠BCD = 90° – 60° = 30°

Now, ∠OBC = ∠BCD + ∠BDC (Exterior angle property)

⇒ 60° = 30° + ∠BDC

⇒ ∠BDC = 30°

∵ ∠BCD = ∠BDC = 30°

∴ BC = BD

**Question: **

The length of tangent from an external point P on a circle with centre O is always less than OP.

**Solution: **True, let PQ be the tangent from the external point P.

Then ∆PQO is always a right angled triangle with OP as the hypotenuse. So, PQ is always less than OP.

**Question: **

If angle between two tangents drawn from a point P to a circle of radius ‘a’ and centre 0 is 90°, then OP = a√2.

**Solution: **

True, let PQ and PR be the tangents

Since ∠P = 90°, so ∠QOR = 90°

Also, OR = OQ = a

∴ PQOR is a square

**Question: **In Fig. 8.20, PA and PB are tangents to the circle drawn from an external point P. CD is the third tangent touching the circle at Q. If PA = 15 cm, find the perimeter of ∆PCD.

**Solution: **∵ PA and PB are tangent from same external point

∴ PA = PB = 15 cm

Now, Perimeter of ∆PCD = PC + CD + DP = PC + CQ + QD + DP

= PC + CA + DB + DP

= PA + PB = 15 + 15 = 30 cm

**Question: **A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre 0 at a point so that OQ = 12 cm. Find the length of PQ.

**Solution:**

We have, ∠OPQ = 90°

OQ = 12 cm and OP = 5 cm

∴ By Pythagoras Theorem

OQ^{2} = OP^{2} + QP^{2}

⇒ 122 = 52 + QP^{2}

⇒ QP^{2} = 144 – 25 = 119

= QP = √119 cm

**Question:** From a point l, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. Find the radius of the circle.

:

**Solution: **Let QT be the tangent and OT be the radius of circle. Therefore

OT ⊥ QT i.e., ∠OTQ = 90°

and OQ = 25 cm and QT = 24 cm

Now, by Pythagoras Theorem, we have

OQ^{2} = QT^{2} + OT^{2}

⇒ 25^{2} = 24^{2} + OT^{2}

⇒ OT^{2} = 25^{2} – 24^{2}

⇒ 625 – 576

OT^{2} – 49

∴ OT = 7 cm

**Question: **In Fig. 8.35, if TP and TQ are the two tangents to a circle with centre 0 so that ∠POQ = 110°, then find ∠PTQ.

**Solution:**

Since TP and TQ are the tangents to the circle with centre O

So, OPIPT and OQ ⊥ QT

⇒ ∠OPT = 90°, ∠OQT = 90° and ∠POQ = 110°

So, in quadrilateral OPTQ, we have

∠POQ + ∠OPT + ∠PTQ + ∠TQO = 360°

⇒ 110° + 90° + ∠PTQ + 90° = 360°

⇒ ∠PTQ + 290° = 360°

∴ ∠PTQ = 360° – 290°

= ∠PTQ = 70°

**Question: **Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

**Solution: **

Let AB be the diameter of the given circle with centre O, and two tangents PQ and LM are drawn at the end of diameter AB respectively.

Р. Now, since the tangent at a point to a circle is perpendicular to the radius through the point of contact.

Therefore, OA ⊥ PQ and OB ⊥ LM

i.e., AB ⊥ PQ and also AB ⊥ LM

⇒ ∠BAQ = ∠ABL (each 90°)

∴ PQ||LM (∵ ∠BAQ and ∠ABL are alternate angles)

**Question: **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 find ∠POA.

**Solution:**

∵ PA and PB are tangents to a circle with centre O,

∴ OA ⊥ AP and OB ⊥ PB

i.e., ∠APB = 80°, ∠OAP = 90°, and ∠OBP = 90°

Now, in quadrilateral OAPB, we have

∠APB + ∠PBO + ∠BOA + ∠OAP = 360°

⇒ 80° + 90° + ∠BOA + 90o = 360°

⇒ 260° + ∠BOA = 360°

∴ ∠BOA = 360° – 260°

⇒ ∠BOA = 100°

Now, in ∆POA and APOB we have

OP = OP (Common)

ОА = ОВ (Radii of the same circle)

∠OAP = ∠OBP = 90°

∴ ∆POA ≅ APOB (RHS congruence condition)

⇒ ∠POA = ∠POB (CPCT)

Now, ∠POA = 12 = ∠BOA = 12 × 100 = 50°

**Question:** In Fig. 8.35, if TP and TQ are the two tangents to a circle with centre 0 so that ∠POQ = 110°, then find ∠PTQ.

**Solution:**

Since TP and TQ are the tangents to the circle with centre O

So, OPIPT and OQ ⊥ QT

⇒ ∠OPT = 90°, ∠OQT = 90° and ∠POQ = 110°

So, in quadrilateral OPTQ, we have

∠POQ + ∠OPT + ∠PTQ + ∠TQO = 360°

⇒ 110° + 90° + ∠PTQ + 90° = 360°

⇒ ∠PTQ + 290° = 360°

∴ ∠PTQ = 360° – 290°

= ∠PTQ = 70°

## Extra Questions for Class 10 Maths Chapter 10 Long Answer Type

**Question: **Prove that the tangent to a circle is perpendicular to the radius through the point of contact.

**Solution:** Given: A circle C(O,r) and a tangent AB at a point P.

To Prove: OP ⊥ AB.

Construction: Take any point l, other than P, on the tangent AB. Join OQ. Suppose OQ meets the circle at R.

Proof: We know that among all line segments joining the point to a point on AB, the shortest one is perpendicular to AB. So, to prove that OP ⊥ AB it is sufficient to prove that OP is shorter than any other segment joining O to any point of AB.

Clearly, OP = OR [Radii of the same circle]

Now, OQ = OR + RQ

⇒ OQ > OR

⇒ OQ > OP [∵OP = OR]

Thus, OP is shorter than any other segment joining O to any point on AB.

Hence, OP ⊥ AB.

**Question: **Prove that the lengths of two tangents drawn from an external point to a circle are equal.

**Solution:**

Given: AP and AQ are two tangents from a point A to a circle C (O, r).

To Prove: AP = AQ

Construction: Join OP, OQ and OA.

Proof: In order to prove that AP = AQ, we shall first prove that ∆OPA ≅ ∆OQA.

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

∴ OP ⊥ AP and OQ ⊥ AQ

⇒ ∠OPA = ∠OQA = 90°

Now, in right triangles OPA and OQA, we have

OP = OQ [Radii of a circle]

∠OPA = ∠OQA [Each 90°]

and OA = OA [Common]

So, by RHS-criterion of congruence, we get

∆OPA ≅ OQA

⇒ AP = AQ [CPCT]

Hence, lengths of two tangents from an external point are equal.

**Question: **Prove that the parallelogram circumscribing a circle is a rhombus.

**Solution: **

Let ABCD be a parallelogram such that its sides touch a circle with centre O.

We know that the tangents to a circle from an exterior point are equal in length.

Therefore, we have

AP = AS [Tangents from A]

BP = BQ [Tangents from B] …. (ii)

CR = CQ [Tangents from C] …. (iii)

And DR = DS [Tangents from D] …. (iv)

Adding (i), (ii), (iii) and (iv), we have

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

AB + CD = AD + BC

AB + AB = BC + BC [∵ ABCD is a parallelogram ∴ AB = CD, BC = DA]

2AB = 2BC ⇒ AB = BC

Thus, AB = BC = CD = AD

Hence, ABCD is a rhombus.

**Question: **In Fig. 8.47, PQ is a chord of length 16 cm, of a circle of radius 10 cm. The tangents at P and Q intersect at a point T. Find the length of TP.

**Solution:** Given: PQ = 16 cm

PO = 10 cm

**Question:** If PQ is a tangent drawn from an external point P to a circle with centre O and QOR is a diameter where length of QOR is 8 cm such that ∠POR = 120°, then find OP and PQ.

**Solution: **Let O be the centre and QOR = 8 cm is diameter of a circle. PQ is tangent such that ∠POR = 120°

**Question: **In Fig. 8.50, O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle.

**Solution:** In right ∆POT

TE = 8 cm

Let PA = AE = x

(Tangents from an external point to a circle are equal)

In right ∆AET

TA^{2} = TE^{2} + EA^{2}

⇒ (12 – x)^{2} = 64 + x^{2}

⇒ 144 + x^{2} – 24x = 64 + x^{2}

⇒ x = 80/24

⇒ x = 3.3 cm

Thus, AB = 6.6 cm