IGCSE · Practice questions · Circle theorems II: centre, segments, cyclic quadrilaterals

Name the theorem.

Extended. Six original Cambridge-style questions on the angle at the centre, angles in the same segment, cyclic quadrilaterals, the alternate segment theorem and the symmetry of chords and tangents.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Which arc? Which theorem?

Centre involved? Angle at centre = 2 × angle at circumference on the same arc.
Two angles on the same arc? Same segment: equal.
Four points on the circle? Opposite angles add to \(180^\circ\).
Tangent and chord? Alternate segment theorem; tangent and radius \(90^\circ\).
Chord or two tangents? Perpendicular from centre bisects; tangents from a point are equal.
Reasons in syllabus words for every step.

Recall

[3 marks]

Complete each statement.

(a) Angles in the same segment are ... (b) The tangents from an external point to a circle are ... (c) Equal chords are ... from the centre.

(a) equal ✓
(b) equal in length ✓
(c) equidistant (the same distance) ✓

Calculation

[3 marks]

A, B, C and D lie on a circle with centre O. C and D are on the major arc AB, and angle ADB \(= 38^\circ\). Find (a) angle ACB (b) angle AOB. Give reasons.

(a) Angle ACB = angle ADB \(= 38^\circ\) (angles in the same segment are equal). ✓
(b) Angle AOB \(= 2 \times 38^\circ = 76^\circ\) (angle at the centre is twice the angle at the circumference). ✓

Calculation

[3 marks]

ABCD is a cyclic quadrilateral. Angle ABC \(= 2x + 10^\circ\) and angle ADC \(= 3x - 5^\circ\). Find \(x\) and both angles.

Opposite angles of a cyclic quadrilateral add to \(180^\circ\): \(5x + 5 = 180\). ✓
\(5x = 175\), so \(x = 35\). ✓
Angle ABC \(= 80^\circ\) and angle ADC \(= 100^\circ\). ✓

Calculation

[3 marks]

A circle has centre O and radius 13 cm. Chord PQ has length 24 cm. Find the distance from O to PQ. A second chord RS is also 24 cm long: how far is it from O?

The perpendicular from O bisects PQ, so the half-chord is 12 cm. ✓
Distance \(= \sqrt{13^2 - 12^2} = \sqrt{25} = 5\) cm. ✓
RS is also 5 cm from O (equal chords are equidistant from the centre). ✓

Proof

[4 marks]

ABCD is a cyclic quadrilateral with centre O inside it. Let angle DAB \(= a\) and angle BCD \(= c\).

Prove that \(a + c = 180^\circ\).

Angle DAB stands on arc BCD, so the angle at the centre on that arc is \(2a\) (angle at centre is twice the angle at circumference). ✓
Angle BCD stands on arc DAB, so the angle at the centre on that arc is \(2c\). ✓
These two centre angles together make the full turn at O: \(2a + 2c = 360^\circ\) (angles at a point). ✓
Therefore \(a + c = 180^\circ\). ✓

Analysis

[6 marks]

A circle has centre O and radius 6 cm. P is a point 10 cm from O. PA and PB are tangents from P, touching the circle at A and B. C is a point on the major arc AB.

(a) Find PA. (b) Write down PB, giving a reason. (c) Find angle APB, to 1 decimal place. (d) Find angle ACB, to 1 decimal place.

(a) Angle OAP \(= 90^\circ\) (tangent and radius), so \(PA = \sqrt{10^2 - 6^2} = \sqrt{64} = 8\) cm. ✓
(b) PB = 8 cm: tangents from an external point are equal. ✓
(c) \(\sin(\text{APO}) = \dfrac{6}{10}\), so angle APO \(= 36.87^\circ\). By symmetry, angle APB \(= 2 \times 36.87 = 73.7^\circ\). ✓
(d) Angle AOB \(= 360 - 90 - 90 - 73.74 = 106.26^\circ\) (angles in quadrilateral OAPB). ✓
Angle ACB \(= 106.26 \div 2 = 53.1^\circ\) (angle at the centre is twice the angle at the circumference). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.