Circle theorems - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 36 exam-style questions on Circle theorems for Edexcel GCSE Maths (Non-calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

36 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

The diagram shows a circle with centre O.
AB is a diameter of the circle and C is a point on the circle.
Angle CAB = 27°.

Work out the size of angle ABC, marked x.

Geometry diagram, marked "Diagram NOT accurately drawn". A circle with centre O; A, B and C lie on it, in that order going anticlockwise. Triangle ABC. O lies on side AB. Angle CAB is marked 27°. Angle ABC is marked x.

Model answer

Angle ACB = 90° (angle in a semicircle)
\(x = 180 - 90 - 27 = 63^\circ\)

Explanation

Because AB is a diameter, the angle at C is 90°. The angles in triangle ABC add up to 180°, so the other two angles add up to 90°.

Mark scheme

1 mark
M1 for angle ACB = 90° or \(180 - 90 - 27\)
correct answer with no working scores full marks
1 mark
A1 for 63°
cao

Question 2

2 marks

The diagram shows a circle with centre O.
A, B and C are points on the circle and AB is a diameter.
The tangent to the circle at B meets the line AC extended at T.
Angle BAC = 35°.

Work out the size of angle CBT, marked x.

Geometry diagram, marked "Diagram NOT accurately drawn". A circle with centre O; A, B and C lie on it, in that order going anticlockwise. Line segments AB, AT and BC are drawn. A straight line touches the circle at B (the tangent at B); it passes through B and T. O lies on AB. C lies on AT. Angle BAC is marked 35°. Angle CBT is marked x.

Model answer

Angle ACB = 90° (angle in a semicircle), so angle ABC \(= 180 - 90 - 35 = 55^\circ\).
Angle ABT = 90° (tangent is perpendicular to the radius OB).
\(x = 90 - 55 = 35^\circ\)

Explanation

The diameter gives a right angle at C, and the tangent at B is perpendicular to the diameter AB. Alternatively, the alternate segment theorem gives angle CBT = angle CAB = 35° straight away.

Mark scheme

1 mark
M1 for angle ABT = 90° and angle ACB = 90° (or angle ABC = 55°), or for stating the alternate segment theorem
correct answer with no working scores full marks
1 mark
A1 for 35°
cao

Question 3

3 marks

The diagram shows a circle with centre O.
A and B are points on the circle.
M is the point on the chord AB such that OM is perpendicular to AB.

Prove that M is the midpoint of AB.

Geometry diagram, marked "Diagram NOT accurately drawn". A circle with centre O; A and B lie on it. Line segments AB, OM, OA and OB are drawn. M lies on AB. Angle OMB is marked as a right angle (90°).

Model answer

Consider triangles OMA and OMB.
OA = OB (radii of the circle)
OM is common to both triangles.
Angle OMA = angle OMB = 90° (given)
So triangles OMA and OMB are congruent (RHS).
Therefore AM = BM, so M is the midpoint of AB.

Explanation

To prove two lengths are equal, show they are corresponding sides of congruent triangles. Here both triangles have a right angle, a hypotenuse that is a radius, and the shared side OM, which is the RHS condition.

Mark scheme

1 mark
M1 for OA = OB (radii) and OM is common to triangles OMA and OMB
1 mark
M1 for angle OMA = angle OMB = 90°, so triangles OMA and OMB are congruent (RHS)
or using Pythagoras: \(AM^2 = OA^2 - OM^2\) and \(BM^2 = OB^2 - OM^2\)
1 mark
C1 for concluding AM = BM, so M is the midpoint of AB, with complete reasons

Question 4

4 marks

The diagram shows four points A, B, C and D on a circle.
AB = AD.
DCE is a straight line.
Angle BCE = 76°.

Work out the size of angle ABD, marked x.
Give reasons for your answer.

Geometry diagram, marked "Diagram NOT accurately drawn". A circle; A, B, C and D lie on it, in that order going anticlockwise. Quadrilateral ABCD. Line segments CE and BD are drawn. Angle BCE is marked 76°. Angle ABD is marked x. AB and AD are each marked with one short dash (equal lengths).

Model answer

Angle BCD \(= 180 - 76 = 104^\circ\) (angles on a straight line add up to 180°)
Angle BAD \(= 180 - 104 = 76^\circ\) (opposite angles of a cyclic quadrilateral add up to 180°)
AB = AD, so triangle ABD is isosceles.
\(x = (180 - 76) \div 2 = 52^\circ\) (base angles of an isosceles triangle are equal)

Explanation

The exterior angle of a cyclic quadrilateral equals the interior opposite angle, so angle BAD = 76°. Triangle ABD is isosceles because AB = AD, so the remaining 104° is shared equally between angles ABD and ADB.

Mark scheme

1 mark
M1 for angle BCD = 104° or angle BAD = 76°
1 mark
M1 for \((180 - 76) \div 2\)
ft their angle BAD
1 mark
A1 for 52°
cao
1 mark
C1 for reasons: angles on a straight line add up to 180°; opposite angles of a cyclic quadrilateral add up to 180°; base angles of an isosceles triangle are equal
accept 'the exterior angle of a cyclic quadrilateral equals the interior opposite angle' in place of the first two

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