Circles — Class 10 Maths practice questions

Question 1

A circular clock face has its centre at point O and a radius of 13 cm. Two straight cracks appear on the clock face during transport: crack PQ and crack RS, both chords of the circle. Crack PQ is 24 cm long and crack RS is 10 cm long. Both cracks are on the same side of the centre O, and the perpendicular distances from O to each crack are to be determined for insurance assessment. The perpendicular from O meets PQ at M and RS at N.

Accepted Answer

**(i)**

By the theorem — **the perpendicular from the centre of a circle to a chord bisects the chord** — OM ⊥ PQ gives:

PM = PQ/2 = 24/2 = 12 cm

In right △OMP, by the Pythagoras theorem:

OM² + PM² = OP²

⟹ OM² + 12² = 13²

⟹ OM² = 169 − 144 = 25

∴ OM = **5 cm**

---

**(ii)**

By the same theorem, ON ⊥ RS gives:

RN = RS/2 = 10/2 = 5 cm

In right △ONR, by the Pythagoras theorem:

ON² + RN² = OR²

⟹ ON² + 5² = 13²

⟹ ON² = 169 − 25 = 144

∴ ON = **12 cm**

---

**(iii)**

Let two equal chords AB and CD each be 24 cm long (AB = CD = 24 cm), with the perpendicular from O meeting AB at P and CD at Q.

By the theorem — the perpendicular from the centre bisects the chord:

AP = AB/2 = 24/2 = 12 cm, and CQ = CD/2 = 24/2 = 12 cm

In right △OAP (OA = radius = 13 cm):

OP² = OA² − AP² = 13² − 12² = 169 − 144 = 25

⟹ OP = 5 cm

In right △OCQ (OC = radius = 13 cm):

OQ² = OC² − CQ² = 13² − 12² = 169 − 144 = 25

⟹ OQ = 5 cm

Since OP = OQ = 5 cm, the two equal chords AB and CD are equidistant from the centre O.

∴ **Equal chords of a circle are equidistant from the centre. Each of AB and CD is at a distance of 5 cm from O.**

Now, for chord EF at perpendicular distance equal to the radius (13 cm) from O:

Let the foot of perpendicular be T. In right △OTE, OT = 13 cm and OE = 13 cm (radius).

(EF/2)² = OE² − OT² = 13² − 13² = 0

⟹ EF/2 = 0 ⟹ EF = 0 cm

∴ **EF has zero length — it degenerates to a point, meaning the line is a tangent to the circle touching it at exactly one point.**

Question 2

A circular garden has its centre at point O and a radius of 10 cm (in a scale model). A straight path AB is drawn as a chord of the circle such that the perpendicular distance from the centre O to the chord AB is 6 cm. A gardener wants to place decorative lights along the chord AB and also along the arc. Another chord CD is drawn parallel to AB at a distance of 8 cm from the centre O. Both chords lie on the same side of the centre.

Accepted Answer

**(i) Length of chord AB**

Let M be the foot of the perpendicular from O to chord AB.

Given: OM = 6 cm, OA = 10 cm (radius).

By the theorem **"The perpendicular from the centre to a chord bisects the chord"**, M is the mid-point of AB.

In △OMA, applying the **Pythagoras Theorem**:

AM² = OA² − OM²

⟹ AM² = 10² − 6² = 100 − 36 = 64

⟹ AM = 8 cm

∴ AB = 2 × AM = 2 × 8 = **16 cm**

---

**(ii) Length of chord CD**

Let N be the foot of the perpendicular from O to chord CD.

Given: ON = 8 cm, OC = 10 cm (radius).

By the theorem **"The perpendicular from the centre to a chord bisects the chord"**, N is the mid-point of CD.

In △ONC, applying the **Pythagoras Theorem**:

CN² = OC² − ON²

⟹ CN² = 10² − 8² = 100 − 64 = 36

⟹ CN = 6 cm

∴ CD = 2 × CN = 2 × 6 = **12 cm**

---

**(iii) Distance between the two parallel chords AB and CD**

Since both chords lie on the **same side** of centre O, the perpendicular distances from O are:

OM = 6 cm (to AB) and ON = 8 cm (to CD), where M and N lie on the same line through O, on the same side.

∴ Distance between the two chords = ON − OM = 8 − 6 = **2 cm**

Question 3

A circular fountain in a city park has its centre at point O and a radius of 15 m. Two straight benches are placed along chords AB and CD of the fountain's boundary. Bench AB is 18 m long and bench CD is 24 m long. The perpendicular distance from the centre O to bench AB is represented as p, and the perpendicular distance from centre O to bench CD is represented as q. A maintenance worker needs to find these distances to plan the underground pipe layout beneath the park.

Accepted Answer

**(i)**

The perpendicular from the centre of a circle to a chord bisects the chord.

For chord AB = 18 m, the perpendicular from O bisects AB.

⟹ Half of AB = 18/2 = 9 m, Radius = 15 m

Applying Pythagoras' Theorem in the right triangle formed:

p² + 9² = 15²

⟹ p² = 225 − 81 = 144

⟹ p = 12 m

∴ The perpendicular distance from centre O to bench AB, **p = 12 m**.

---

**(ii)**

For chord CD = 24 m, the perpendicular from O bisects CD.

⟹ Half of CD = 24/2 = 12 m, Radius = 15 m

Applying Pythagoras' Theorem in the right triangle formed:

q² + 12² = 15²

⟹ q² = 225 − 144 = 81

⟹ q = 9 m

∴ The perpendicular distance from centre O to bench CD, **q = 9 m**.

---

**(iii)**

A third bench EF is placed at a perpendicular distance of 12 m from O (equal to p = 12 m).

The perpendicular from O bisects EF. Let half the length of EF = d.

Applying Pythagoras' Theorem:

d² + 12² = 15²

⟹ d² = 225 − 144 = 81

⟹ d = 9 m

⟹ Length of EF = 2 × 9 = **18 m**

**Verification:** By the theorem — Chords equidistant from the centre of a circle are equal in length.

Since bench AB and bench EF are both at perpendicular distance 12 m from O, EF = AB = 18 m. ✓

∴ The length of bench EF = **18 m**, and EF = AB, verified.

Question 4

A decorative showpiece in an art gallery is made by placing a solid cone on top of a solid hemisphere, both sharing the same circular base of radius 7 cm. The total height of the combined showpiece (from the bottom of the hemisphere to the tip of the cone) is 16 cm. The showpiece is to be painted, and the artist also wants to know the total volume of material used to make it. (Use π = 22/7)

Accepted Answer

**(i) Height of the cone** Radius of hemisphere = 7 cm ⟹ height of hemispherical part = 7 cm. Height of cone = Total height − height of hemisphere ⟹ Height of cone = 16 − 7 = **9 cm** --- **(ii) Volume of the hemisphere** Formula: Volume of hemisphere = (2/3)πr³ ⟹ V = (2/3) × (22/7) × 7³ ⟹ V = (2/3) × (22/7) × 343 ⟹ V = (2/3) × 22 × 49 ⟹ V = 2156/3 ∴ Volume of hemisphere = **2156/3 cm³ ≈ 718.67 cm³** --- **(iii) Total volume of the showpiece** Given: r = 7 cm, height of cone h = 9 cm. **Volume of cone:** Formula: V_cone = (1/3)πr²h ⟹ V_cone = (1/3) × (22/7) × 7² × 9 ⟹ V_cone = (1/3) × (22/7) × 49 × 9 ⟹ V_cone = (1/3) × 22 × 7 × 9 ⟹ V_cone = (1/3) × 1386 = 462 cm³ **Volume of hemisphere** (from part ii) = 2156/3 cm³ **Total volume:** V_total = V_cone + V_hemisphere ⟹ V_total = 462 + 2156/3 ⟹ V_total = 1386/3 + 2156/3 ⟹ V_total = 3542/3 ∴ Total volume of material used = **3542/3 cm³ ≈ 1180.67 cm³**

Question 5

A tangent is drawn from an external point P to a circle with centre O and radius 12 cm. If the distance OP = 37 cm, what is the length of the tangent from P to the circle?

Accepted Answer

Option (A) is correct.

**Explanation:** By the theorem that the tangent at any point of a circle is perpendicular to the radius through that point, OT ⊥ PT, where T is the point of tangency. Thus △OTP is right-angled at T with OP as hypotenuse. ⟹ PT = √(OP² − OT²) = √(37² − 12²) = √(1369 − 144) = √1225 = 35 cm.

Question 6

Two tangents MN and MP are drawn from an external point M to a circle with centre O and radius 5 cm. If the length of each tangent is 12 cm, then the perimeter of △ONM is:

Accepted Answer

Option (A) is correct.

**Explanation:** Since a tangent is perpendicular to the radius at the point of contact, ON ⊥ MN, so △ONM is right-angled at N. By the Pythagoras theorem, MO = √(ON² + MN²) = √(5² + 12²) = √(25 + 144) = √169 = 13 cm. ∴ Perimeter of △ONM = ON + NM + MO = 5 + 12 + 13 = **30 cm**.

Question 7

In the given figure, PQ is a tangent drawn from an external point P to a circle with centre O. If OP = 10 cm and the radius OQ = 6 cm, then the length of the tangent PQ is:

Accepted Answer

Option (A) is correct.

Explanation: Since PQ is a tangent to the circle at Q, by the theorem "radius drawn to the point of tangency is perpendicular to the tangent," OQ ⊥ PQ, making △OQP right-angled at Q. By Pythagoras' theorem, PQ² = OP² − OQ² = 10² − 6² = 100 − 36 = 64 ⟹ PQ = √64 = 8 cm.

Question 8

A point P is at a distance of 25 cm from the centre O of a circle of radius 7 cm. The length of the tangent drawn from P to the circle is:

Accepted Answer

Option (A) is correct.

**Explanation:** The tangent to a circle is perpendicular to the radius at the point of tangency, so △OAP is right-angled at A. By the Pythagoras theorem: PA² = OP² − OA² = 25² − 7² = 625 − 49 = 576 ⟹ PA = √576 = **24 cm**.

Question 9

From an external point T, a tangent TP is drawn to a circle with centre O such that OT = 10 cm and ∠OTP = 60°. What is the length of the tangent TP?

Accepted Answer

Option (A) is correct.

**Explanation:** The radius to the point of tangency is perpendicular to the tangent (radius ⊥ tangent), so ∠OPT = 90°, making △OTP a right triangle with the right angle at P. In △OTP, OT = 10 cm and ∠OTP = 60°. Applying the cosine ratio: cos 60° = TP/OT ⟹ 1/2 = TP/10 ⟹ TP = 5 cm.

Question 10

A tangent is drawn from an external point C to a circle with centre O and radius 20 cm. If the distance OC = 29 cm, then the length of the tangent from C to the circle is:

Accepted Answer

Option (A) is correct.

Explanation: The tangent to a circle is perpendicular to the radius at the point of tangency, so △OAC is right-angled at A (where A is the point of tangency). By the Pythagoras Theorem: CA² = OC² − OA² = 29² − 20² = 841 − 400 = 441 ⟹ CA = √441 = 21 cm.

Question 11

In a circle with centre O, a tangent TQ is drawn from an external point T such that OT = 6 cm and ∠OTQ = 30°. What is the length of the tangent TQ?

Accepted Answer

Option (B) is correct.

**Explanation:** The radius OQ is perpendicular to tangent TQ at the point of tangency (radius ⊥ tangent), so ∠OQT = 90°. In right △OQT, cos(∠OTQ) = TQ/OT ⟹ cos 30° = TQ/6 ⟹ TQ = 6 × (√3/2) ∴ TQ = 3√3 cm.

Question 12

A tangent is drawn from an external point to a circle of radius 5 cm. If the distance from the external point to the centre of the circle is 13 cm, what is the length of the tangent?

Accepted Answer

Option (A) is correct.

Explanation: Since the tangent is perpendicular to the radius at the point of tangency, △OTP is right-angled at T. By the Pythagoras Theorem, PT² = OP² − OT² = 13² − 5² = 169 − 25 = 144 ⟹ PT = √144 = 12 cm.

Question 13

Assertion (A): The tangent at any point of a circle is perpendicular to the radius.
Reason (R): A tangent to a circle touches it at exactly one point.

Accepted Answer

Option (b) is correct.

**Explanation:** Assertion (A) states that the tangent at any point of a circle is perpendicular to the radius through that point — this is a standard theorem (Theorem: The tangent at any point of a circle is perpendicular to the radius through the point of contact) and is **true**. Reason (R) states that a tangent touches a circle at exactly one point — this is the definition of a tangent and is also **true**. However, R does not logically explain A; the perpendicularity of the tangent and radius is established by a contradiction argument (any other line through the point of contact intersects the circle at two points, making it a secant, so the perpendicular from the centre is the shortest distance), which is independent of R merely stating that the tangent meets the circle at one point. ∴ Both A and R are true but R is not the correct explanation of A.

Question 14

From an external point P, a tangent PQ is drawn to a circle with centre O and radius 6 cm. If ∠OPQ = 30°, what is the length of the tangent PQ?

Accepted Answer

Option (A) is correct.

**Explanation:** The radius OQ is perpendicular to tangent PQ (radius ⊥ tangent at point of contact), so △OPQ is right-angled at Q. In △OPQ, tan(∠OPQ) = OQ/PQ ⟹ tan 30° = 6/PQ ⟹ (1/√3) = 6/PQ ⟹ PQ = 6√3 cm.

Question 15

From an external point G, two tangents GE and GF are drawn to a circle with centre O and radius 7 cm. If OG = 25 cm, then the length (GE + GF) is:

Accepted Answer

Option (A) is correct.

**Explanation:** Since the radius is perpendicular to the tangent at the point of tangency, △OEG is right-angled at E. By the Pythagoras theorem, GE = √(OG² − OE²) = √(25² − 7²) = √(625 − 49) = √576 = 24 cm. By the property that tangents drawn from an external point to a circle are equal in length, GE = GF = 24 cm. ∴ GE + GF = 24 + 24 = **48 cm**.

Question 16

From an external point Q, two tangents QA and QB are drawn to a circle of radius 6 cm with centre O. If ∠AQB = 60°, then the length of each tangent QA is:

Accepted Answer

Option (A) is correct.

**Explanation:** Since tangents from an external point are equal, QA = QB, so OQ bisects ∠AQB, giving ∠AQO = 30°. By the theorem "radius is perpendicular to the tangent at the point of contact," OA ⊥ QA, forming right △OAQ. Applying tan∠AQO = OA/QA ⟹ tan 30° = 6/QA ⟹ 1/√3 = 6/QA ⟹ QA = 6√3 cm.

Question 17

From an external point B, a tangent is drawn to a circle with centre O and radius 6 cm. If the distance OB = 10 cm, what is the length of the tangent from B to the circle?

Accepted Answer

Option (B) is correct.

Explanation: The radius to the point of tangency is perpendicular to the tangent (radius ⊥ tangent). By the Pythagoras theorem in the right triangle formed, tangent² = OB² − r² = 10² − 6² = 100 − 36 = 64 ⟹ tangent length = √64 = 8 cm.

Question 18

From an external point D, a tangent is drawn to a circle with centre O and radius 9 cm. If the distance OD = 41 cm, what is the length of the tangent from D to the circle?

Accepted Answer

Option (A) is correct.

Explanation: The radius OT is perpendicular to the tangent DT at the point of contact T, so △OTD is right-angled at T. By the Pythagoras Theorem, DT² = OD² − OT² = 41² − 9² = 1681 − 81 = 1600 ⟹ DT = √1600 = 40 cm.

Question 19

Two tangents inclined at an angle of 120° are drawn from an external point to a circle of radius 4 cm. What is the length of each tangent?

Accepted Answer

Option (B) is correct.

**Explanation:** The radius is perpendicular to the tangent at the point of contact. The angle between the two tangents is 120°, so the angle between each tangent and the line joining the external point to the centre is 120°/2 = 60°. In the right triangle formed by the radius (r = 4 cm), the tangent length (l), and the line to the centre: tan 60° = r/l ⟹ √3 = 4/l ⟹ l = 4/√3 cm.

Question 20

A tangent LM is drawn from an external point L to a circle with centre O and radius 15 cm. If the distance OL = 17 cm, what is the length of the tangent LM?

Accepted Answer

Option (A) is correct.

**Explanation:** The tangent to a circle is perpendicular to the radius at the point of tangency, so OM ⊥ LM and △OML is right-angled at M. By the Pythagoras theorem, LM² = OL² − OM² = 17² − 15² = 289 − 225 = 64 ⟹ LM = √64 = **8 cm**.

Circles: Class 10 Maths Practice Questions

Want unlimited practice on Circles?