Extended. Four angle theorems (angle at the centre, same segment, cyclic quadrilateral, alternate segment) and three symmetry properties of chords and tangents complete the circle toolkit.
Angle at the centre \(= 2 \times\) angle at the circumference (same arc). Angles in the same segment are equal. Opposite angles of a cyclic quadrilateral add to \(180^\circ\). The angle between a tangent and a chord equals the angle in the alternate segment. Tangents from one point are equal, and the perpendicular from the centre bisects a chord.
Fig. 1: arc AB subtends 100° at the centre O and 50° at C on the circumference: the angle at the centre is twice the angle at the circumference. Drawn to scale
Section 01
The angle at the centre.
In the angle-at-centre tab, drag C around the major arc and watch angle ACB stay at exactly half of angle AOB. Then move A and B apart until O falls outside triangle ACB: the rule still holds. Afterwards, open the same-segment tab and see two angles standing on the same arc stay equal.
Angle at the centre: the angle subtended by an arc at the centre is twice the angle it subtends at any point on the remaining part of the circumference (Fig. 1).
Angles in the same segment are equal: angles at the circumference standing on the same arc (on the same side of the chord) are equal, because each is half the same angle at the centre.
Reflex case: if the point on the circumference is on the minor arc, the angle at the centre to use is the REFLEX angle.
The semicircle theorem is the special case where the angle at the centre is \(180^\circ\).
Worked example Centre and circumference
A, B and C lie on a circle with centre O. Angle AOB \(= 124^\circ\) and C is on the major arc. Find angle ACB. Another point D lies on the major arc. Find angle ADB.
Angle ACB \(= 124^\circ \div 2 = 62^\circ\) (angle at the centre is twice the angle at the circumference).
D and C are in the same segment, standing on arc AB.
Angle ADB = angle ACB \(= 62^\circ\) (angles in the same segment are equal).
A and B lie on a circle with centre O, and angle AOB \(= 130^\circ\). Point E lies on the minor arc AB. Find angle AEB.
E is on the minor arc, so angle AEB stands on the major arc AB.
The angle at the centre on that side is the reflex angle: \(360^\circ - 130^\circ = 230^\circ\).
Angle AEB \(= 230^\circ \div 2 = 115^\circ\) (angle at the centre is twice the angle at the circumference).
Answer: \(115^\circ\)
Examiner trap
Halve the angle at the centre to get the circumference angle, do not double it. And check which arc each angle stands on: if the angle at the circumference is obtuse, the matching centre angle is reflex.
Section 02
Cyclic quadrilaterals.
In the cyclic quadrilateral tab, drag any of the four corners around the circle. The readouts show that opposite angles always total 180°. Pull one corner off the circle and the rule breaks.
Cyclic quadrilateral: all four vertices lie on a circle.
Theorem: opposite angles of a cyclic quadrilateral add to \(180^\circ\) (they are supplementary).
Consequence: an exterior angle of a cyclic quadrilateral equals the interior angle at the opposite vertex.
Worked example Opposite angles
ABCD is a cyclic quadrilateral. Angle DAB \(= 3x\) and angle BCD \(= x + 40^\circ\). Find \(x\) and angle DAB.
A and C are opposite vertices, so \(3x + x + 40 = 180\) (opposite angles of a cyclic quadrilateral).
ABCD is a cyclic quadrilateral and side AB is extended beyond B to E. Angle ADC \(= 77^\circ\). Find angle CBE.
Angle ABC \(= 180^\circ - 77^\circ = 103^\circ\) (opposite angles of a cyclic quadrilateral). ABE is a straight line, so angle CBE \(= 180^\circ - 103^\circ = 77^\circ\), equal to angle ADC.
Examiner trap
Only OPPOSITE angles add to \(180^\circ\). Adjacent angles of a cyclic quadrilateral have no special rule, and a quadrilateral whose vertices are not all on the circle has no such rule at all.
Stage 1 · Learn
Check your understanding
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The angle subtended by an arc at the centre is ... the angle it subtends at the circumference.
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Opposite angles of a cyclic quadrilateral
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Two tangents are drawn from a point P outside a circle. What is true about their lengths from P to the points of contact?
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A perpendicular is drawn from the centre of a circle to a chord. What does it do to the chord?
Section 03
The alternate segment.
In the alternate segment tab, drag the point C around the major arc. The angle between the tangent and the chord stays equal to angle ACB, the angle in the opposite (alternate) segment.
Theorem: the angle between a tangent and a chord at the point of contact equals the angle subtended by that chord in the alternate segment.
Finding the right angle pair: the angle between tangent and chord AB at A is on one side of AB; the matching angle is at a point C on the circle on the OTHER side of AB, in triangle ABC.
Worked example Tangent and chord
TA is a tangent to a circle at A. The chord AB makes an angle of \(58^\circ\) with the tangent. C is a point on the circle in the alternate segment, and angle CAB \(= 47^\circ\). Find angles ACB and ABC.
The equal angle is in the ALTERNATE segment, across the chord from the tangent angle. Matching it with an angle in the same segment as the tangent angle is the usual error.
Section 04
Chords and tangents: symmetry.
In the chords tab, slide a chord across the circle. The perpendicular from O always hits the chord at its midpoint, and two chords of the same length always sit the same distance from O.
The perpendicular bisector of a chord passes through the centre. Equivalently, the perpendicular from the centre to a chord bisects it.
Equal chords are equidistant from the centre (and chords the same distance from the centre are equal).
Tangents from an external point are equal in length. If PA and PB are tangents from P, then PA = PB, triangle PAB is isosceles, and OAPB is a kite with right angles at A and B.
Worked example Distance of a chord from the centre
A circle has radius 10 cm. A chord AB is 16 cm long. Find the distance from the centre O to the chord.
The perpendicular from O to AB meets it at the midpoint M (perpendicular from the centre bisects the chord), so AM = 8 cm.
Triangle OMA has a right angle at M, with hypotenuse OA = 10 cm.
\(OM = \sqrt{10^2 - 8^2} = \sqrt{36} = 6\) cm.
Answer: 6 cm
Worked example Two tangents
PA and PB are tangents from P to a circle with centre O, touching at A and B. Angle APB \(= 48^\circ\). Find angle AOB and angle PAB.
Angles OAP and OBP are \(90^\circ\) (tangent and radius).
Angles in quadrilateral OAPB add to \(360^\circ\): angle AOB \(= 360 - 90 - 90 - 48 = 132^\circ\).
PA = PB (tangents from an external point are equal), so triangle PAB is isosceles.
In the two-tangents example, C is a point on the major arc AB. Find angle ACB in two different ways.
Angle at the centre: \(132^\circ \div 2 = 66^\circ\). Alternate segment: the angle between tangent PA and chord AB is \(66^\circ\), so angle ACB \(= 66^\circ\). Both agree.
Examiner trap
The right angles in the tangent kite are at A and B (where the tangents touch), not at P or O. Angle AOB and angle APB are supplementary, not equal.
Stage 2 · Exam
Exam-style questions
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A, B and C lie on a circle with centre O, with C on the major arc. Angle AOB \(= 140^\circ\). Find angle ACB.
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ABCD is a cyclic quadrilateral with angle ABC \(= 82^\circ\). Find angle ADC.
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TA is a tangent to a circle at A, and the chord AB makes an angle of \(64^\circ\) with TA. C is on the circle in the alternate segment. Find angle ACB.
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PA and PB are tangents from P to a circle with centre O. Angle APB \(= 40^\circ\). Find angle OAB.
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The circle theorems, complete.
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