A circle is just every point sitting at the SAME distance from one fixed centre. From that single idea, a whole world of equal chords, perfect angles and clever theorems unfolds → master it and geometry feels like magic.
Chord & centre
The perpendicular from the centre to a chord cuts it exactly in half.
Equal chords
Equal chords are equally far from the centre, and vice-versa.
Angle at centre
The centre angle is double the angle on the remaining arc.
Cyclic quad
Opposite angles of a cyclic quadrilateral add up to 180°.
1. Basic terms you must know
A circle is the set of all points in a plane that are at a fixed distance (the radius, r) from a fixed point (the centre, O). A line segment joining the centre to any point on the circle is a radius; a segment joining two points on the circle is a chord. The longest chord, which passes through the centre, is the diameter (d = 2r). A piece of the circle between two points is an arc — the smaller one is the minor arc and the larger one the major arc. The region between a chord and its arc is a segment, while the region between two radii and an arc is a sector. The whole boundary length is the circumference. Points all lying on one circle are called concyclic.
2. Equal chords and the angles they make
Theorem: Equal chords of a circle subtend equal angles at the centre. If two chords AB and CD are equal, then ∠AOB = ∠COD. The proof uses congruent triangles: OA = OC and OB = OD (all radii), and AB = CD (given), so △AOB ≅ △COD by SSS, giving equal angles. The converse is also true: if two chords subtend equal angles at the centre, the chords are equal. This pair of results lets us swap freely between "equal chords" and "equal central angles."
3. The perpendicular from the centre to a chord
Theorem: The perpendicular drawn from the centre of a circle to a chord bisects the chord. If OM ⊥ AB, then AM = MB. The proof joins OA and OB (radii); triangles OMA and OMB are right-angled, share OM, and have equal hypotenuses OA = OB, so by the RHS rule they are congruent, giving AM = MB. The converse says: the line drawn from the centre to the midpoint of a chord is perpendicular to that chord. So "from the centre" + "bisects" automatically means "perpendicular," and "from the centre" + "perpendicular" automatically means "bisects."
4. Circle through three points
Through one point, infinitely many circles can pass. Through two points, infinitely many circles can pass, with their centres lying on the perpendicular bisector of the segment joining them. But through three non-collinear points, exactly one circle passes. To find its centre, draw the perpendicular bisectors of any two of the three joining segments; the single point where they meet is equidistant from all three points and is the centre. This is exactly how the circumcircle of a triangle is constructed — three collinear points cannot lie on a circle.
5. Equal chords and their distance from the centre
Theorem: Equal chords of a circle are equidistant from the centre. Conversely, chords that are equidistant from the centre are equal in length. Distance is always measured along the perpendicular. So a chord nearer the centre is longer, and the diameter (distance 0 from the centre) is the longest chord of all. As a chord moves away from the centre it gets shorter, shrinking to a single point at the circle’s edge.
6. Angle subtended by an arc
Central-angle theorem: The angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the remaining part of the circle. In symbols, ∠POQ = 2 × ∠PAQ, where A is on the major arc. A direct consequence is the angle in a semicircle: an angle in a semicircle is a right angle (90°), because the arc’s central angle is the straight angle 180°, and half of 180° is 90°.
7. Angles in the same segment
Theorem: Angles in the same segment of a circle are equal. If two points A and B both lie on the major arc above a chord PQ, then ∠PAQ = ∠PBQ, because each equals half of the same central angle ∠POQ. The converse is a powerful test: if a line segment joins two points and subtends equal angles at two other points lying on the same side of it, then all four points are concyclic (they lie on one circle).
8. Cyclic quadrilaterals
A quadrilateral whose four vertices all lie on a circle is a cyclic quadrilateral. Theorem: The sum of either pair of opposite angles of a cyclic quadrilateral is 180° (they are supplementary). The converse is equally useful: if the sum of a pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic. Also, the exterior angle of a cyclic quadrilateral equals the interior opposite angle. These facts solve a huge fraction of exam problems on circles.
- Diameter d = 2r; circumference C = 2πr.
- Perpendicular from centre → bisects the chord (and the converse).
- Equal chords ⇔ equal central angles ⇔ equal distance from centre.
- ∠ at centre = 2 × ∠ at remaining circle on the same arc.
- Angle in a semicircle = 90°.
- Angles in the same segment are equal.
- Cyclic quad: opposite angles sum to 180°.
- Exactly one circle passes through 3 non-collinear points.
A chord of length 24 cm is drawn in a circle of radius 13 cm. Find the distance of the chord from the centre.
- Let O be the centre and AB the chord with AB = 24 cm. Drop OM ⊥ AB.
- The perpendicular from the centre bisects the chord, so AM = MB = 24 ÷ 2 = 12 cm.
- In right triangle OMA, OA = 13 cm (radius) is the hypotenuse and AM = 12 cm.
- By Pythagoras: OM² = OA² − AM² = 13² − 12² = 169 − 144 = 25.
- So OM = √25 = 5 cm.
In a circle with centre O, an arc PQ subtends an angle of 130° at the centre. Point A lies on the major arc. Find ∠PAQ. Also find ∠PBQ if B is the reflex-side point such that PBQA is a cyclic quadrilateral.
- The angle at the centre is double the angle at any point on the remaining (major) arc: ∠POQ = 2 × ∠PAQ.
- So 130° = 2 × ∠PAQ, which gives ∠PAQ = 130° ÷ 2 = 65°.
- Now PBQA is a cyclic quadrilateral, so opposite angles are supplementary: ∠PAQ + ∠PBQ = 180°.
- Therefore ∠PBQ = 180° − 65° = 115°.
"Centre is the boss, it doubles the angle." The central angle is always twice the angle on the rim — the boss (centre) always gets the bigger number. And remember POMD: Perpendicular from the centre → bisects; midpoint line → perpendicular — the two halves of one idea, working both ways.
The biggest mistake is forgetting to take HALF the chord when using Pythagoras — you must use the half-length (AM), not the full chord (AB), because the perpendicular bisects it. Also, the "double angle" rule only works when both angles stand on the same arc; always identify which arc your point lies on before doubling or halving.
Q1. Two equal chords AB and CD of a circle intersect inside the circle. Prove that the line joining the point of intersection to the centre makes equal angles with the two chords.
Answer: Let the chords meet at P and let O be the centre. Draw OM ⊥ AB and ON ⊥ CD. Since AB and CD are equal chords, they are equidistant from the centre, so OM = ON. In right triangles OMP and ONP: OM = ON (just shown), OP = OP (common hypotenuse), and each has a right angle. By the RHS congruence rule, △OMP ≅ △ONP. Hence ∠OPM = ∠OPN, which means OP makes equal angles with the two chords. Proved.
Q2. A chord AB and a diameter of a circle of radius 10 cm are such that the chord is 6 cm from the centre. Find the length of the chord.
Answer: Let O be the centre and OM ⊥ AB with OM = 6 cm; the radius OA = 10 cm. The perpendicular from the centre bisects the chord, so AM = MB. In right triangle OMA, AM² = OA² − OM² = 10² − 6² = 100 − 36 = 64, so AM = √64 = 8 cm. Therefore the full chord AB = 2 × AM = 2 × 8 = 16 cm.
Q3. In a cyclic quadrilateral ABCD, ∠A = 70° and ∠B = 100°. Find ∠C and ∠D.
Answer: In a cyclic quadrilateral, opposite angles are supplementary. ∠A and ∠C are opposite, so ∠C = 180° − ∠A = 180° − 70° = 110°. ∠B and ∠D are opposite, so ∠D = 180° − ∠B = 180° − 100° = 80°. Check: 70 + 100 + 110 + 80 = 360°, which confirms the four angles of the quadrilateral. So ∠C = 110° and ∠D = 80°.
Q4. AB is a diameter of a circle and C is a point on the circle. If ∠BAC = 35°, find ∠ABC.
Answer: Since AB is a diameter, the angle in the semicircle ∠ACB is a right angle, so ∠ACB = 90°. In triangle ACB, the three angles add up to 180°: ∠BAC + ∠ACB + ∠ABC = 180°. Substituting, 35° + 90° + ∠ABC = 180°, so ∠ABC = 180° − 125° = 55°. Therefore ∠ABC = 55°.
- ✅ A circle is all points at distance r from the centre; chord, arc, segment and sector are its parts.
- ✅ The perpendicular from the centre bisects a chord; equal chords are equidistant from the centre.
- ✅ Central angle = 2 × angle on the remaining arc; semicircle angle = 90°; same-segment angles are equal.
- ✅ A cyclic quadrilateral has opposite angles summing to 180°, and exactly one circle passes through 3 non-collinear points.
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