1. Chapter at a Glance
On a giant wheel at a fair you go up and down and round and round — but always at the same distance from the centre. That is exactly what a circle is. Wheels, bangles, the rim of a well, the paths of planets (nearly) and kolam and rangoli designs all use circles. In this chapter we study the circle carefully and prove its beautiful properties: about chords, arcs, angles at the centre and on the circle, and quadrilaterals whose corners lie on a circle.
- Parts of a circle: radius, chord, diameter, arc, segment, sector.
- Equal chords and the angles they subtend at the centre.
- Perpendicular from the centre to a chord.
- One and only one circle through three non-collinear points.
- Equal chords are equidistant from the centre.
- Angle at the centre is double the angle on the circle; angle in a semicircle.
- Angles in the same segment; cyclic quadrilaterals.
2. Parts of a Circle
| Term | Meaning |
|---|---|
| Circle | All points in a plane at a fixed distance (radius r) from a fixed point (centre O) |
| Chord | A segment joining two points on the circle |
| Diameter | A chord through the centre; the longest chord; d = 2r |
| Arc | A part of the circle between two points; minor (smaller) and major (larger) |
| Segment | Region between a chord and its arc |
| Sector | Region between two radii and the arc |
| Concentric circles | Circles with the same centre, different radii |
| Congruent circles | Circles with equal radii |
3. Chords and Angles at the Centre
Equal chords of a circle subtend equal angles at the centre.
Proof: If AB = CD, then in △AOB and △COD: OA = OC, OB = OD (radii), AB = CD. By SSS, the triangles are congruent, so ∠AOB = ∠COD. ∎
Converse: If two chords subtend equal angles at the centre, they are equal (by SAS).
These also hold for congruent circles.
4. Perpendicular from the Centre to a Chord
The perpendicular from the centre to a chord bisects the chord.
Proof: Let OM ⊥ AB. In right triangles OMA and OMB: OA = OB (radii), OM common. By RHS, △OMA ≅ △OMB, so AM = BM. ∎
Converse: The line from the centre to the midpoint of a chord is perpendicular to it.
A chord of a circle of radius 13 cm is 5 cm from the centre. Find the length of the chord.
Solution: Half-chord = √(13² − 5²) = √(169 − 25) = √144 = 12 cm. Chord = 24 cm.
5. Circle Through Three Points
- Through one point: infinitely many circles.
- Through two points: infinitely many circles (centres on the perpendicular bisector of the segment).
- Through three collinear points: no circle.
- Through three non-collinear points: exactly one circle.
There is one and only one circle through three non-collinear points. Its centre is where the perpendicular bisectors of AB and BC meet (the circumcentre of △ABC).
6. Equal Chords and Distance from the Centre
Equal chords of a circle are equidistant from the centre. Conversely, chords equidistant from the centre are equal.
Also, a longer chord is nearer to the centre. The diameter (distance 0) is the longest chord.
7. Angles Subtended by an Arc
The angle subtended by an arc at the centre is double the angle it subtends at any point on the remaining part of the circle. ∠AOB = 2∠ACB.
Proof idea: Join CO and extend it to D. △OAC is isosceles (OA = OC), so ∠OAC = ∠OCA, and the exterior angle ∠AOD = 2∠OCA. Similarly ∠BOD = 2∠OCB. Adding gives ∠AOB = 2∠ACB. ∎
- Angles in the same segment are equal. All points C on the same arc see AB at the same angle.
- Angle in a semicircle is a right angle. If AB is a diameter, ∠AOB = 180°, so ∠ACB = 90°.
∠AOB = 140° at the centre. Find ∠ACB where C is on the major arc.
Solution: ∠ACB = 140° ÷ 2 = 70°.
8. Cyclic Quadrilaterals
A quadrilateral whose four vertices lie on a circle is cyclic.
The opposite angles of a cyclic quadrilateral add up to 180°. Conversely, if a pair of opposite angles of a quadrilateral is supplementary, it is cyclic.
Proof idea: ∠A and ∠C stand on arcs that together make the whole circle (360° at the centre). Each is half its centre angle, so ∠A + ∠C = ½ × 360° = 180°.
Exterior angle property: an exterior angle of a cyclic quadrilateral equals the interior opposite angle.
Every rectangle and every isosceles trapezium is cyclic; a parallelogram is cyclic only if it is a rectangle.
ABCD is cyclic with ∠A = (2x + 10)° and ∠C = (3x − 5)°. Find x and ∠A.
Solution: 5x + 5 = 180 → x = 35. ∠A = 80°, ∠C = 100°.
9. Solved Problems (Board Pattern)
Two parallel chords of a circle of radius 10 cm are 12 cm and 16 cm long, on the same side of the centre. Find the distance between them.
Solution: Distances from O: √(100 − 36) = 8 and √(100 − 64) = 6. Same side → 2 cm apart (opposite sides would give 14 cm).
AB is a diameter and C is on the circle with ∠CAB = 35°. Find ∠CBA.
Solution: ∠ACB = 90°, so ∠CBA = 180° − 90° − 35° = 55°.
Points P and Q lie on the same arc of chord AB, and ∠APB = 48°. Find ∠AQB and the angle AB subtends at the centre.
Solution: ∠AQB = 48° (same segment); ∠AOB = 96°.
10. Master Formula Table
| Result | Statement |
|---|---|
| Diameter | d = 2r (longest chord) |
| Chord, radius, distance | (chord/2)² + d² = r² |
| Centre vs circle angle | ∠AOB = 2∠ACB |
| Angle in semicircle | 90° |
| Same segment | ∠ACB = ∠ADB |
| Cyclic quadrilateral | ∠A + ∠C = ∠B + ∠D = 180° |
| Exterior angle (cyclic) | = interior opposite angle |
11. Master Summary Table
| Theorem | Proof tool | Exam focus |
|---|---|---|
| Equal chords ↔ equal centre angles | SSS / SAS | Prove chords equal |
| Perpendicular bisects chord | RHS | Chord length problems |
| Unique circle through 3 points | Perpendicular bisectors | Circumcentre |
| Equal chords equidistant | RHS | Parallel chord distances |
| Centre angle = 2 × circle angle | Isosceles + exterior angle | Angle chasing |
| Cyclic quadrilateral | Centre-angle theorem | Find unknown angles |
Chart — half-chord length in a circle of radius 10 cm, by distance from centre:
12. Mind Map
- Equal ↔ equal angles
- ⊥ bisects
- Equal chords equidistant
- Longer = nearer
- Centre = 2 × circle
- Semicircle 90°
- Opposite = 180°
- Exterior = interior opposite
13. Tips and Tricks
- In chord problems, always draw the perpendicular from O and a radius to the chord's end — you get a right triangle.
- Look for a diameter: it immediately gives a 90° angle.
- Radii are equal, so triangles with two radii as sides are isosceles — use equal base angles.
- Using "centre angle = 2 × circle angle" when the circle point is on the same arc as AB — it must be on the remaining arc.
- Assuming a quadrilateral is cyclic without checking its vertices lie on the circle.
- For parallel chords, forgetting the "opposite sides of the centre" case.
14. Important One-liners
- The diameter is the longest chord of a circle.
- The perpendicular from the centre to a chord bisects it.
- Exactly one circle passes through three non-collinear points.
- Equal chords are equidistant from the centre.
- The angle in a semicircle is 90°.
- Opposite angles of a cyclic quadrilateral are supplementary.
15. Exam-pattern Questions with Detailed Answers
How many circles pass through three collinear points?
- 0
- 1
- 2
- Infinitely many
Show answer
Collinear points cannot all lie on a circle.
Answer: (A) 0
In a cyclic quadrilateral, ∠B = 72°. Find ∠D.
Show answer
180° − 72° = 108°.
Answer: 108°
An arc subtends 50° at a point on the circle. What does it subtend at the centre?
Show answer
Double: 100°.
Answer: 100°
A chord of length 16 cm is 6 cm from the centre. Find the radius.
Show answer
r² = 8² + 6² = 100 → r = 10 cm.
Answer: 10 cm
A, B, C are on a circle with centre O, and ∠OAB = 25°. Find ∠ACB (C on the major arc).
Show answer
OA = OB, so ∠OBA = 25° and ∠AOB = 130°. ∠ACB = 65°.
Answer: 65°
ABCD is cyclic. ∠DBC = 55° and ∠BAC = 45°. Find ∠BCD.
Show answer
∠BDC = ∠BAC = 45° (same segment, chord BC). In △BCD: ∠BCD = 180° − 55° − 45° = 80°.
Answer: 80°
An exterior angle of a cyclic quadrilateral at B is 110°. Find the interior angle at D.
Show answer
Exterior angle at B equals interior opposite angle D = 110°.
Answer: 110°
Two chords AB = 10 cm and CD = 24 cm are parallel and on opposite sides of the centre of a circle of radius 13 cm. Find the distance between them.
Show answer
Distances: √(169 − 25) = 12, √(169 − 144) = 5. Opposite sides: 17 cm.
Answer: 17 cm
Prove that a cyclic parallelogram is a rectangle.
Show answer
In a parallelogram, ∠A = ∠C. Being cyclic, ∠A + ∠C = 180°. So 2∠A = 180° → ∠A = 90°. A parallelogram with a right angle is a rectangle. ∎
Answer: Proof as shown
Prove that the circle drawn with any side of a rhombus as diameter passes through the point where the diagonals meet.
Show answer
The diagonals of a rhombus meet at O at 90°. For side AB, ∠AOB = 90°. An angle of 90° subtended by AB means O lies on the circle with AB as diameter (converse of the angle-in-a-semicircle theorem). ∎
Answer: Proof as shown
In a circle of radius 5 cm, AB and AC are equal chords of 6 cm. Find the length of BC.
Show answer
Since AB = AC, OA is the perpendicular bisector of BC; let it meet BC at M. Let OM = x, BM = y. In △OMB: x² + y² = 25. In △AMB: (5 − x)² + y² = 36. Subtract: (5 − x)² − x² = 11 → 25 − 10x = 11 → x = 1.4. y² = 25 − 1.96 = 23.04 → y = 4.8. BC = 9.6 cm.
Answer: 9.6 cm
ABCD is cyclic and AB is a diameter. If ∠ADC = 130°, find ∠BAC.
Show answer
∠ABC = 180° − 130° = 50° (cyclic). ∠ACB = 90° (semicircle). In △ABC: ∠BAC = 180° − 90° − 50° = 40°.
Answer: 40°
16. Quick Revision Summary
- Equal chords ↔ equal centre angles ↔ equal distances from the centre.
- Perpendicular from the centre bisects a chord: (chord/2)² + d² = r².
- One circle through three non-collinear points.
- Centre angle = 2 × angle on the circle; same-segment angles equal; semicircle angle 90°.
- Cyclic quadrilateral: opposite angles sum to 180°.