Circles — Practice

Question 1

A chord of length 16 cm is at perpendicular distance 6 cm from the centre. Find the radius.

Accepted Answer

Step 1 — Drop the perpendicular from centre O to chord AB at midpoint M.
  This forms a right triangle OMA with OA = r (radius), OM = 6 (distance), AM = 8 (half-chord).

Step 2 — Apply Pythagoras.
  OA² = OM² + AM²
  r² = 6² + 8² = 36 + 64 = 100.
  r = 10.

✦ Answer: Radius = 10 cm.

Question 2

An arc subtends an angle of 100° at the centre. What angle does it subtend at a point on the major arc?

Accepted Answer

Step 1 — Apply Theorem 4.
  Angle at centre = 2 × angle at any point on the major arc.

Step 2 — Solve.
  Inscribed angle = central angle / 2 = 100° / 2 = 50°.

✦ Answer: 50°.

Question 3

In a circle, AB is a diameter and P is a point on the circle. What is ∠APB?

Accepted Answer

Step 1 — Apply Thales' Theorem (Corollary 2).
  The angle in a semicircle (subtended by a diameter) is 90°.

✦ Answer: ∠APB = 90°.

This is one of the oldest theorems in geometry — attributed to Thales of Miletus circa 600 BCE.

Question 4

In cyclic quadrilateral ABCD, ∠A = 80°. Find ∠C.

Accepted Answer

Step 1 — Apply Cyclic Quadrilateral Theorem.
  Opposite angles of a cyclic quadrilateral sum to 180°.
  ∠A + ∠C = 180°.

Step 2 — Solve.
  80° + ∠C = 180° → ∠C = 100°.

✦ Answer: ∠C = 100°.

Question 5

In a circle, points A, B, C, D lie on the same arc. If ∠ACB = 40°, find ∠ADB.

Accepted Answer

Step 1 — Recall: angles in the same segment are equal.
  ∠ACB and ∠ADB both subtend chord AB and both vertices C, D are on the same arc.

Step 2 — So ∠ADB = ∠ACB = 40°.

✦ Answer: ∠ADB = 40°.

Question 6

In a circle of radius 13 cm, a chord is at distance 5 cm from the centre. Find the length of the chord.

Accepted Answer

Step 1 — Setup the right triangle OMA.
  OA = 13 (radius), OM = 5 (distance from centre to chord).

Step 2 — Find half-chord AM by Pythagoras.
  AM² = OA² − OM² = 169 − 25 = 144.
  AM = 12.

Step 3 — Chord length AB = 2 × AM = 24.

✦ Answer: Chord length = 24 cm.

Question 7

Two chords of a circle, AB = 16 cm and CD = 12 cm. The perpendicular from the centre to AB is 6 cm. Find the perpendicular distance from the centre to CD. (Radius is the same circle.)

Accepted Answer

Step 1 — Use AB to find the radius.
  AB = 16, half-chord = 8. Perpendicular distance from centre = 6.
  r² = 6² + 8² = 36 + 64 = 100 → r = 10.

Step 2 — For chord CD = 12, half-chord = 6.
  Let d = perpendicular distance from centre to CD.
  r² = d² + 6²
  100 = d² + 36 → d² = 64 → d = 8.

✦ Answer: Perpendicular distance from centre to CD = 8 cm.

Observation: longer chord is CLOSER to the centre (AB closer than CD, since shorter chord is farther).

Question 8

In a circle, an arc subtends an angle of 130° at the centre. Find the angles subtended at any point on (a) the major arc, (b) the minor arc.

Accepted Answer

Step 1 — Major arc.
  By Theorem 4, angle at any point on the MAJOR arc = (1/2) × central angle = 130° / 2 = 65°.

Step 2 — Minor arc.
  A point on the MINOR arc subtends an angle related to the REFLEX of the central angle.
  Reflex angle = 360° − 130° = 230°.
  Inscribed angle on minor arc = 230° / 2 = 115°.

Step 3 — Verify by cyclic-quadrilateral logic.
  The angle on major arc (65°) + angle on minor arc (115°) = 180°. ✓ (They subtend supplementary inscribed angles — a special case of the cyclic theorem.)

✦ Answer: 65° on the major arc, 115° on the minor arc.

Question 9

In cyclic quadrilateral ABCD, ∠A : ∠B = 2 : 3. If ∠C = 70°, find ∠A, ∠B, ∠D.

Accepted Answer

Step 1 — Use Cyclic Theorem for opposite angles.
  ∠A + ∠C = 180° → ∠A = 180° − 70° = 110°.

Step 2 — Find ∠B from the ratio.
  ∠A : ∠B = 2 : 3 → 110 : ∠B = 2 : 3.
  ∠B = (3/2) × 110 = 165°.

Wait — let me check. If ∠A = 110° from the cyclic property, but ∠A : ∠B = 2 : 3 gives ∠B = 165°. Then ∠B + ∠D = 180° → ∠D = 180 − 165 = 15°.

Step 3 — Verify all four angles sum to 360°.
  110 + 165 + 70 + 15 = 360. ✓

✦ Answer: ∠A = 110°, ∠B = 165°, ∠C = 70° (given), ∠D = 15°.

Note: the problem's ratio combined with the given ∠C forces ∠B to be quite large. Worth verifying the angles are valid (each < 360° — yes).

Question 10

Two equal chords AB and CD of a circle intersect at point P inside the circle. Prove that PA = PD (or equivalently, the segments from P to corresponding endpoints are equal).

Accepted Answer

Step 1 — Equal chords are equidistant from the centre (Theorem 2).
  Let OM ⊥ AB at M, ON ⊥ CD at N. AB = CD ⇒ OM = ON.

Step 2 — In △OMP and △ONP.
  OM = ON (Step 1).
  ∠OMP = ∠ONP = 90°.
  OP = OP (common).
  By RHS, △OMP ≅ △ONP.

Step 3 — By CPCT, MP = NP.

Step 4 — Since M is midpoint of AB, AM = MB. Since N is midpoint of CD, CN = ND.
  So PA = AM − MP and PD = ND − NP (depending on which side of M and N point P lies; specific arrangement may vary).
  Given the symmetry MP = NP and AM = ND, we get PA = PD.

✦ Answer: PA = PD. ∎

This is a classic 'use perpendicular distance to relate chords' style problem.

Question 11

Prove Theorem 4 for the case where the centre lies INSIDE △APB (where ∠APB is the inscribed angle).

Accepted Answer

Step 1 — Setup. Centre O inside △APB. Draw radii OA, OB, OP. Extend PO to meet the circle at Q (on the major arc, opposite side from P).

Step 2 — In △OPA, OA = OP = r (radii).
  So △OPA is isosceles → ∠OAP = ∠OPA. (Isosceles theorem.)

Step 3 — The exterior angle of △OPA at O is ∠AOQ.
  By the Exterior Angle Theorem of a triangle:
  ∠AOQ = ∠OAP + ∠OPA = 2 ∠OPA.

Step 4 — Similarly in △OPB.
  OB = OP → △OPB isosceles → ∠OBP = ∠OPB.
  Exterior angle: ∠BOQ = ∠OBP + ∠OPB = 2 ∠OPB.

Step 5 — Add the two exterior angles at O.
  ∠AOQ + ∠BOQ = 2 ∠OPA + 2 ∠OPB = 2 (∠OPA + ∠OPB).

Step 6 — Recognise: ∠AOQ + ∠BOQ = ∠AOB (since O is inside ∠APB, the two angles at O sum to the central angle).
  And ∠OPA + ∠OPB = ∠APB.

Step 7 — Therefore ∠AOB = 2 ∠APB. ∎

✦ Answer: ∠AOB = 2 ∠APB.

Note: similar proofs handle the other cases (O outside △APB or on a side).

Question 12

Prove: if a parallelogram is cyclic, it must be a rectangle.

Accepted Answer

Step 1 — Setup. Let ABCD be a parallelogram that's also cyclic.

Step 2 — Apply parallelogram property.
  Adjacent angles are supplementary: ∠A + ∠B = 180°.

Step 3 — Apply cyclic property.
  Opposite angles sum to 180°: ∠A + ∠C = 180°.

Step 4 — Also from parallelogram, ∠A = ∠C (opposite angles equal).

Step 5 — Substitute ∠C = ∠A into Step 3:
  ∠A + ∠A = 180° → 2∠A = 180° → ∠A = 90°.

Step 6 — So all four angles are 90° (∠A = ∠C = 90°; ∠B = ∠D = 90° by similar argument).
  A parallelogram with all 90° angles is a rectangle.

✦ Answer: A cyclic parallelogram is necessarily a rectangle. ∎

Elegant interaction between two chapter theorems.

Practice — Circles

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