Circles: Important Questions and Answers for Class 10

Important questions on Circles for Class 10 are provided in this article to help students prepare effectively for the CBSE board exams. These questions are carefully designed to align with the latest CBSE syllabus and exam pattern, covering key concepts such as tangents to a circle and their properties. Since Circles is an important chapter in Class 10 Maths, practising these questions can help students improve conceptual understanding, accuracy, and problem-solving skills. Our subject experts have prepared well-structured questions with detailed solutions to support complete revision and boost confidence before the examination.

Table of Contents

• Key Concepts on Circles for Class 10
• Important Questions on Tangent Length
• Important Questions on Theorem & Proofs
• Important Questions on Concentric Circles
• Important Questions on Angles & Tangent-Radius
• HOTS & Previous Year Board Questions

Key Concepts on Circles for Class 10

Theorem 1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

In other words: if PQ is a tangent to a circle at point A, and OA is the radius, then OA ⊥ PQ and ∠OAP = ∠OAQ = 90°.

Theorem 2: The lengths of tangents drawn from an external point to a circle are equal.

In other words: if PA and PB are tangents from external point P to a circle with centre O, then PA = PB.

Important Questions on Tangent Length

Q1: 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. Find the radius of the circle.

Solution: Let the tangent touch the circle at point P. Since the radius OP is perpendicular to the tangent PQ at the point of contact:

∠OPQ = 90° (Theorem 1: tangent ⊥ radius)

△OPQ is right-angled at P.

OQ² = OP² + PQ² 

⟹ 25² = OP² + 24²

625 = OP² + 576 

⟹ OP² = 49 

⟹ OP = 7 cm

Radius of the circle = 7 cm

Q 2: A point P is 13 cm from the centre of a circle. The length of the tangent drawn from P to the circle is 12 cm. Find the radius of the circle.

Solution: Given, OP = 13 cm, tangent length PT = 12 cm

By Pythagoras: OP² = OT² + PT² 

⟹ 169 = OT² + 144

OT² = 25 

⟹ OT = 5 cm

Radius = 5 cm

Q3: If the angle between two tangents drawn from an external point P to a circle of radius a and centre O is 60°, find the length of OP.

Solution: Let the two tangents from P touch the circle at A and B. 

Since ∠APB = 60°, the line OP bisects this angle, so ∠OPA = 30°.

In right △OAP: ∠OAP = 90°, OA = a (radius), ∠OPA = 30°

sin(∠OPA) = OA/OP 

⟹ sin 30° = a/OP

⟹ 1/2 = a/OP 

⟹ OP = 2a

Important Questions on Theorem & Proofs

Q 4: Prove that the lengths of tangents drawn from an external point to a circle are equal.

Solution: Given: A circle with centre O. P is an external point. 

PA and PB are tangents touching the circle at A and B respectively.

To Prove: PA = PB

Construction: Join OA, OB, and OP.

Proof: OA ⊥ PA and OB ⊥ PB (Theorem 1: radius ⊥ tangent) 

⟹ ∠OAP = ∠OBP = 90°

In △OAP and △OBP: 

OA = OB (radii of the same circle) 

OP = OP (common hypotenuse) 

∠OAP = ∠OBP = 90°

By RHS congruence: △OAP ≅ △OBP

∴ PA = PB (by CPCT)

Hence proved: tangents from an external point are equal in length.

Q 5: Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Solution: Given: A circle with centre O. AB is a diameter. PQ is a tangent at A and RS is a tangent at B.

To Prove: PQ ∥ RS

OA ⊥ PQ (radius ⊥ tangent) 

⟹ ∠OAP = ∠OAQ = 90°

OB ⊥ RS (radius ⊥ tangent) 

⟹ ∠OBR = ∠OBS = 90°

∠OAQ = ∠OBR = 90° (these are alternate interior angles formed by transversal AB cutting lines PQ and RS)

Since alternate interior angles are equal, PQ ∥ RS.

Hence proved: tangents at the ends of a diameter are parallel.

Q6: A quadrilateral ABCD circumscribes a circle. Prove that AB + CD = AD + BC.

Solution: Given: Quadrilateral ABCD with a circle touching sides AB, BC, CD, and DA at points P, Q, R, and S, respectively.

To Prove: AB + CD = AD + BC

Since tangents from an external point are equal: From A: AP = AS From B: BP = BQ From C: CQ = CR From D: DR = DS

Adding all four equalities: AP + BP + CQ + DR = AS + BQ + CR + DS

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

⟹ AB + CD = AD + BC

Hence proved: AB + CD = AD + BC

Important Questions on Concentric Circles

Q7: Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Solution: Given: Two concentric circles with centre O, radii 5 cm (large) and 3 cm (small). Chord AB of the larger circle is tangent to the smaller circle at point P.

Since AB is a tangent to the smaller circle at P: OP ⊥ AB (Theorem 1) 

And OP = 3 cm (radius of smaller circle), OA = 5 cm (radius of larger circle)

In right △OPA: OA² = OP² + AP² 

⟹ 25 = 9 + AP² 

⟹ AP² = 16 

⟹ AP = 4 cm

Since OP ⊥ AB, the perpendicular from centre bisects the chord: AP = PB = 4 cm

AB = 2 × AP = 2 × 4 = 8 cm

Length of the chord = 8 cm

Important Questions on Angles & Tangent-Radius

Q8: Two tangents, TP and TQ are drawn from an external point T to a circle with centre O. Prove that ∠PTQ = 2∠OPQ.

Solution: 

Given: Tangents TP and TQ from external point T. Let ∠PTQ = θ.

Since TP = TQ (equal tangents), △TPQ is isosceles. ∠TPQ = ∠TQP = ½(180° − θ) = 90° − θ/2

∠OPT = 90° (radius ⊥ tangent)

∠OPQ = ∠OPT − ∠TPQ = 90° − (90° − θ/2) = θ/2

∴ ∠PTQ = θ = 2 × (θ/2) = 2∠OPQ

Hence proved: ∠PTQ = 2∠OPQ

Q9: From external point P, two tangents, PT and PS are drawn to a circle with centre O and radius r. If OP = 2r, show that ∠OTS = ∠OST = 30°.

Solution: 

In △OTP: OT ⊥ TP (Theorem 1), OT = r

OP = 2r sin(∠OPT) = OT/OP = r/2r = 1/2 

⟹ ∠OPT = 30°

By symmetry (PT = PS, OT = OS): ∠OPS = 30° ∴ ∠TPS = ∠OPT + ∠OPS = 60°

In △TPS: PT = PS

so ∠PTS = ∠PST = (180° − 60°)/2 = 60°

∠OTS + ∠PTS = 90° (since OT ⊥ PT) 

∠OTS = 90° − 60° = 30°

Since OT = OS (radii), ∠OTS = ∠OST = 30°

Hence proved: ∠OTS = ∠OST = 30°

Q10: Two tangents, RQ and RP are drawn from external point R to a circle with centre O. If ∠PRQ = 120°, prove that OR = PR + RQ.

Solution: OR bisects ∠PRQ (since △OPR ≅ △OQR), ∴ ∠PRO = ∠QRO = 60°

In right △OPR (∠OPR = 90°): cos 60° = PR/OR 

⟹ 1/2 = PR/OR 

⟹OR = 2PR

Similarly, OR = 2QR (by symmetry)

Since PR = QR (equal tangents): OR = 2PR = PR + PR = PR + QR (since PR = QR)

Hence proved: OR = PR + RQ

Q11: PQ is a chord of a circle with centre O. PT is a tangent. If ∠QPT = 60°, find ∠PRQ (where R is a point on the major arc).

Solution: ∠QPT = 60° (given)

Using the tangent-chord theorem:

The angle between a tangent and a chord is equal to the angle in the alternate segment.

Therefore, ∠QPT = ∠PRQ.

Since ∠QPT = 60°

We get ∠PRQ = 60°.

HOTS & Previous Year Board Questions

HOTS 1: Prove that the parallelogram circumscribing a circle is a rhombus.

Solution:

Given: Parallelogram ABCD circumscribes a circle.

To Prove: ABCD is a rhombus (all sides are equal).

Since ABCD circumscribes a circle, by the property proved earlier, AB + CD = AD + BC ...(i)

Since ABCD is a parallelogram, AB = CD and AD = BC ...(ii)

Substituting (ii) in (i): AB + AB = AD + AD 

⟹ 2AB = 2AD 

⟹AB = AD

Since AB = CD, AD = BC, and AB = AD

All four sides are equal ⟹ ABCD is a rhombus.

Hence proved: the parallelogram is a rhombus.

 

HOTS 2: PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at point T. Find the length of TP.

Solution:

Given: Circle with centre O, radius 5 cm. Chord PQ = 8 cm. Tangents at P and Q meet at T.

Let M be the midpoint of PQ. Then OM ⊥ PQ and PM = 4 cm. 

OM² = OP² − PM² = 25 − 16 = 9 

⟹ OM = 3 cm

Let OT = d and TP = t. 

In △OPT: ∠OPT = 90° (tangent ⊥ radius) 

OT² = OP² + PT² 

⟹ d² = 25 + t² ...(i)

Since TM ⊥ PQ (T lies on the perpendicular bisector of PQ by symmetry),

TM = TO − OM = d − 3 (if T is outside and O, M, T are collinear)

Also, in △TPM: TP² = TM² + PM², so t² = (d − 3)² + 16. 

⟹ t² = d² − 6d + 9 + 16 

⟹ t² = d² − 6d + 25 ...(ii)

From (i): d² = 25 + t². 

Sub into (ii): t² = (25 + t²) − 6d + 25 

⟹ 0 = 50 − 6d 

⟹ d = 25/3

From (i): t² = (25/3)² − 25 = 625/9 − 225/9 = 400/9 

⟹ t = 20/3 cm

∴ Length of TP = 20/3 cm ≈ 6.67 cm

HOTS 3: Prove that the tangent drawn at the midpoint of an arc of a circle is parallel to the chord joining the endpoints of that arc.

Solution: 

Let AB be a chord of a circle with centre O, and let P be the midpoint of the arc AB. A tangent is drawn at P meeting the circle at P.

We need to prove that the tangent at P is parallel to the chord AB.

Since P is the midpoint of arc AB, arc AP = arc PB

So the angles subtended at the centre are equal:

∠AOP = ∠POB

Hence, OP bisects the central angle ∠AOB.

The tangent at point P is perpendicular to the radius OP:

PT⊥OP

We compare angles made by the chord AB and radius lines.

In △AOB, since OP is the angle bisector:

∠AOP = ∠POB

So OP is the axis of symmetry of chord AB. This implies:

Points A and B are symmetric about OP

Therefore, line AB⊥OP

So we get:

AB⊥OP

We now have:

Tangent PT⊥OP

Chord AB⊥OP

Lines perpendicular to the same line are parallel.

So, PT∥AB

Hence proved: the tangent at the midpoint of the arc is parallel to the chord AB.