Practise 28 exam-style questions on Circle theorems for Edexcel GCSE Maths (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.
28 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. A, B and C are points on the circle. Angle AOB = 148°.
Work out the size of angle ACB, marked x. Give a reason for your answer.
Model answer
x = 148° ÷ 2 = 74° The angle at the centre is twice the angle at the circumference.
Explanation
Angles AOB and ACB both stand on the arc AB. The angle at the centre is double the angle at the circumference, so halve 148°. In reasons, use the standard wording of the circle theorem - "angle at the centre is twice the angle at the circumference".
Mark scheme
1 mark
B1 for 74°
cao
1 mark
C1 for a correct reason, e.g. the angle at the centre is twice the angle at the circumference
accept "angle subtended at the centre is double the angle subtended at the circumference"; do not accept "angle at the centre is double" without reference to the circumference
Question 2
2 marks
The diagram shows a circle with centre O. A, B, C and D are points on the circle.
Use the fact that the angle at the centre is twice the angle at the circumference to prove that angle ACB = angle ADB.
Model answer
Angle AOB = 2 × angle ACB (angle at the centre is twice the angle at the circumference) Angle AOB = 2 × angle ADB (angle at the centre is twice the angle at the circumference) So 2 × angle ACB = 2 × angle ADB Therefore angle ACB = angle ADB.
Explanation
Both angles at C and D stand on the same arc AB as the angle AOB at the centre, so each is half of angle AOB. Two things that are each half of the same angle must be equal - this proves that angles in the same segment are equal.
Mark scheme
1 mark
M1 for angle AOB = 2 × angle ACB and angle AOB = 2 × angle ADB
both statements needed
1 mark
C1 for completing the proof, e.g. 2 × angle ACB = 2 × angle ADB, so angle ACB = angle ADB
Question 3
3 marks
The diagram shows a circle with centre O. P is a point outside the circle. PA and PB are tangents to the circle at A and B.
Prove that PA = PB.
Model answer
Join OA, OB and OP. OA = OB (radii of the same circle) Angle OAP = angle OBP = 90° (a tangent is perpendicular to the radius) OP is common to both triangles. So triangles OAP and OBP are congruent (RHS). Therefore PA = PB.
Explanation
To prove two lengths are equal, show that they are corresponding sides of congruent triangles. The radii and the right angles between tangents and radii give you RHS congruence. Every statement in a proof needs a reason, such as "radii" or "tangent perpendicular to radius".
Mark scheme
1 mark
M1 for OA = OB (radii) and OP is common to triangles OAP and OBP
1 mark
M1 for angle OAP = angle OBP = 90° because a tangent is perpendicular to the radius
1 mark
C1 for a complete proof, e.g. triangles OAP and OBP are congruent (RHS), so PA = PB
or uses Pythagoras: \(PA^2 = OP^2 - OA^2 = OP^2 - OB^2 = PB^2\); all reasons must be given
Question 4
4 marks
The diagram shows a circle with centre O. A, B and C are points on the circle. The straight line through A and T is the tangent to the circle at A. Angle TAC = 3x° and angle AOC = (5x + 12)°.
Work out the value of x. Give a reason for each stage of your working.
Model answer
Angle ABC = angle TAC = 3x° (the angle between a tangent and a chord is equal to the angle in the alternate segment) Angle AOC = 2 × angle ABC = 6x° (the angle at the centre is twice the angle at the circumference) 6x = 5x + 12 x = 12
Explanation
Link the tangent angle to an angle in the circle with the alternate segment theorem, then link that to the angle at the centre. This gives two expressions for angle AOC to set equal. Check: x = 12 gives angle TAC = 36° and angle AOC = 72°, which is 2 × 36°.
Mark scheme
1 mark
M1 for angle ABC = 3x°
1 mark
M1 for angle AOC = 2 × 3x° = 6x° and 6x = 5x + 12
oe
1 mark
A1 for x = 12
cao
1 mark
C1 for both reasons: the angle between a tangent and a chord is equal to the angle in the alternate segment, and the angle at the centre is twice the angle at the circumference
or uses angle OAT = 90° (tangent perpendicular to radius) and the isosceles triangle OAC: angle OAC = 90 − 3x, so 5x + 12 + 2(90 − 3x) = 180
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