Circle chords and angle properties

Learning outcome

Calculate chord lengths and circle angles using perpendicular distances, intercepted arcs and vertex positions.

Concepts and assumptions

Let a circle have centre O and radius r > 0. A chord joins distinct circle points; a diameter passes through O and has length 2r. Arc points exclude endpoints. Minor arcs measure below 180°; major arcs measure above 180°.

The perpendicular from O bisects a chord. For chord length c and perpendicular distance d, Pythagoras gives:

(c ÷ 2)² + d² = r².

In one circle or equal-radius circles, equal chords have equal perpendicular distances from the centres and equal non-reflex central angles. For a fixed radius, a chord nearer the centre is longer.

For an inscribed angle ∠APB, A, P and B are distinct circle points. It equals half the angular measure of arc AB not containing P. A central angle subtending that same arc is twice the inscribed angle; for a major intercepted arc, use the reflex central angle.

Inscribed angles on the same chord are equal for vertices on the same side of it; opposite sides give supplementary angles because the intercepted arcs total 360°. A diameter subtends 90° at every other point on the circle.

For a quadrilateral whose four vertices lie consecutively on one circle, opposite interior angles total 180°. This cyclic condition is essential.

Worked examples

Example 1 — Chord length. A circle has radius 13 cm. The perpendicular from O meets chord AB at M, with OM = 5 cm. Since AM = MB, right triangle OMA gives AM² = 13² − 5² = 144. Hence AM = 12 cm and AB = 24 cm.

Example 2 — One chord, different arcs. The minor central angle ∠AOB is 104°. Points P and Q lie on the major arc AB, while R lies on the minor arc AB. Therefore ∠APB = ∠AQB = 104° ÷ 2 = 52°. But ∠ARB intercepts the major arc, so ∠ARB = (360° − 104°) ÷ 2 = 128°.

Example 3 — Diameter and triangle. A, B and C lie on a circle; AB is a diameter and ∠BAC = 34°. The angle opposite the diameter is ∠ACB = 90°. Triangle ABC then gives ∠ABC = 180° − 90° − 34° = 56°.

Common mistakes

Do not halve the minor central angle when the vertex lies on the minor arc. Do not use “same chord, equal angles” without checking the segment. The chord-distance formula uses half the chord, not the whole chord.

Practice questions

1. A circle has radius 10 cm. A chord is at perpendicular distance 6 cm from its centre. Find its length.
2. The minor central angle ∠AOB is 138°. P lies on the major arc AB. Find ∠APB.
3. P and Q are distinct points on the major arc AB. If ∠APB = 47°, find ∠AQB and the minor arc AB’s angular measure.
4. A, B, C, D lie consecutively on a circle. Given ∠ABC = 112° and ∠BAD = 73°, find ∠ADC and ∠BCD.

Worked solutions

1. Half-chord² = 10² − 6² = 64, so half-chord = 8 cm. The chord is 2 × 8 = 16 cm.
2. P intercepts the minor arc AB. The inscribed angle is half its central angle: ∠APB = 138° ÷ 2 = 69°.
3. Both vertices occupy the same segment, so ∠AQB = 47°. The intercepted minor arc measures twice that angle: 2 × 47° = 94°.
4. The quadrilateral is cyclic. Opposite angles supplement: ∠ADC = 180° − 112° = 68° and ∠BCD = 180° − 73° = 107°.