OCR GCSE Mathematics: Circle theorems and proofs (J560) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 8.05 (Circles) of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026). Question 1 uses the 8.05a vocabulary, which both tiers need. The other questions use 8.05b to 8.05h, which are in the Higher tier column only, so each is labelled (Higher tier only). Draw a diagram for each question and write a reason beside every angle.

Read the study guide first if a theorem is new, and use the revision notes to recall the reasons. See also the OCR GCSE Mathematics hub, the printable checklist and the free diagnostics.

Questions

1. P and Q are two points on a circle with centre O. The line PQ does not pass through O. Name each of the following.

(a) The line PQ. [1] (b) The region between PQ and the shorter curved path from P to Q. [1] (c) The shorter curved path from P to Q. [1] (d) The region bounded by OP, OQ and the shorter curved path from P to Q. [1]

2. (Higher tier only) A, B and C lie on a circle with centre O, with C on the major arc AB. Angle AOB = 154°.

(a) Find angle ACB, giving a reason. [2] (b) D is a point on the minor arc AB. Find angle ADB, giving a reason. [2]

3. (Higher tier only) PQ is a diameter of a circle and has length 20 cm. R is a point on the circle. Angle PQR = (4y - 3)° and angle RPQ = (y + 8)°.

(a) Find y and the size of each angle. [3] (b) Find the length PR, to 3 significant figures. [2]

4. (Higher tier only) A, B, C and D lie on a circle. Chords AC and BD cross at X. Angle ACB = 35° and angle CAD = 48°.

(a) Find angle ADB, giving a reason. [2] (b) Find angle CBD, giving a reason. [2] (c) Find angle BXC. [2]

5. (Higher tier only) O is the centre of a circle. AB is a chord and M is a point on AB such that OM is perpendicular to AB. Prove that AM = MB. [3]

6. (Higher tier only) A circle has centre O and radius 7.3 cm.

(a) Chord AB is 4.8 cm from O. Find the length of AB. [3] (b) Chord CD is parallel to AB, on the opposite side of O, and has length 9.6 cm. Find the distance between AB and CD. [3]

7. (Higher tier only) The tangent PT touches a circle at P. The centre is O and OP = 4.5 cm. Angle OTP = 28°.

(a) Find angle POT, giving a reason. [2] (b) Find the length TP, to 3 significant figures. [2]

8. (Higher tier only) A straight tangent touches a circle at A. B and C are points on the circle. The tangent makes an angle of 49° with chord AB on one side and 63° with chord AC on the other side. Find the three angles of triangle ABC, giving a reason for each. [4]

9. (Higher tier only) ABCD is a cyclic quadrilateral. Angle DAB = (4x - 6)°, angle BCD = (2x + 30)° and angle ABC = 71°.

(a) Find x. [3] (b) Find angle ADC, giving a reason. [2]

10. (Higher tier only) ABCD is a cyclic quadrilateral with centre O. Prove that angle BAD + angle BCD = 180°. [4]

11. (Higher tier only) The tangents to a circle at A and at B meet at T. The centre is O and angle ATB = 56°. C is a point on the major arc AB and D is a point on the minor arc AB.

(a) Find angle AOB, giving reasons. [2] (b) Find angle ACB, giving a reason. [2] (c) Use the alternate segment theorem to find angle TAB. [2] (d) Find angle ADB, giving a reason. [2]

Answers

1. (a) Chord [1] (b) Segment (the minor segment) [1] (c) Arc (the minor arc) [1] (d) Sector (the minor sector) [1] Examiner insight: Learn each term as a short definition and match the description to the shape. A segment touches a chord and a sector touches the centre, which is the easiest way to tell them apart.

2. (a) Angle ACB = 154° ÷ 2 [1] = 77°, because the angle at the centre is twice the angle at the circumference [1] (b) Angle ADB = 180° - 77° [1] = 103°, because opposite angles of a cyclic quadrilateral ACBD add up to 180° [1] Examiner insight: Write the reason beside the angle. A correct angle with no named theorem may not earn full credit when the question says “giving a reason”.

3. (a) Angle PRQ = 90°, because the angle in a semicircle is a right angle [1]; (4y - 3) + (y + 8) + 90 = 180, so 5y + 95 = 180 [1]; y = 17, so angle PQR = 65° and angle RPQ = 25° [1] (b) PR is adjacent to the 25° angle and PQ is the hypotenuse, so PR = 20 × cos 25° [1] = 18.1 cm [1] Examiner insight: Spot the semicircle first. Without the right angle at R you cannot form the equation or choose cosine. Round only at the end, to the accuracy asked.

4. (a) Angle ADB = 35°, because angles ACB and ADB are in the same segment, both standing on chord AB [1] [1] (b) Angle CBD = 48°, because angles CAD and CBD are in the same segment, both standing on chord CD [1] [1] (c) In triangle XBC, angle XCB = 35° and angle XBC = 48° [1], so angle BXC = 180° - 35° - 48° = 97° [1] Examiner insight: Name the chord that both angles stand on. Naming it shows you have chosen the right pair and avoids the common slip of pairing angles on different chords.

5. OA = OB because both are radii [1]. Angle OMA = angle OMB = 90°, and OM is common to both triangles, so triangles OMA and OMB are congruent (right angle, hypotenuse, side) [1]. So AM = MB (corresponding sides of congruent triangles) [1] Examiner insight: A proof needs a reason for every line and a closing statement of the result. Say which congruence condition you use and list the three equal parts.

6. (a) The perpendicular from the centre bisects the chord, so triangle OMA is right-angled at the midpoint M [1]; AM² = 7.3² - 4.8² = 53.29 - 23.04 = 30.25, so AM = 5.5 [1]; AB = 11 cm [1] (b) Half of CD = 4.8 cm, so the distance from O to CD is √(7.3² - 4.8²) = 5.5 cm [1]; the distance from O to AB is 4.8 cm [1]; the chords are on opposite sides of O, so the distance between them is 4.8 + 5.5 = 10.3 cm [1] Examiner insight: Show the right-angled triangle and the half-chord. After you get a length, check it against the diagram: a half-chord must be shorter than the radius.

7. (a) Angle OPT = 90°, because the radius is perpendicular to the tangent [1]; angle POT = 180° - 90° - 28° = 62° [1] (b) tan 28° = 4.5 / TP [1], so TP = 4.5 ÷ tan 28° = 8.46 cm [1] Examiner insight: Start with the tangent-radius right angle. Choose the trigonometric ratio from the sides you know, and keep the unrounded value in your working until the final line.

8. Angle ACB = 49°, because of the alternate segment theorem with the angle between tangent and AB [1]. Angle ABC = 63°, by the alternate segment theorem with the angle between tangent and AC [1]. Angle BAC = 180° - 49° - 63° [1] = 68°, because angles on a straight line add up to 180° [1] Examiner insight: The alternate segment theorem links the tangent-chord angle to the angle in the segment on the opposite side of the chord. Check your answer by confirming the three angles sum to 180°.

9. (a) Opposite angles of a cyclic quadrilateral add up to 180°, so (4x - 6) + (2x + 30) = 180 [1]; 6x + 24 = 180 [1]; x = 26 [1] (b) Angle ADC = 180° - 71° [1] = 109°, because angle ADC is opposite angle ABC in a cyclic quadrilateral [1] Examiner insight: Identify which angles are opposite before you write the equation. Substitute x back to check, here angle DAB = 98° and angle BCD = 82°, which total 180°.

10. Let angle BAD = a and angle BCD = c. The arc BCD subtends angle 2a at the centre [1]. The arc BAD subtends angle 2c at the centre [1]. These two angles at O fill the full turn around O, so 2a + 2c = 360° [1]. Dividing by 2, a + c = 180° [1] Examiner insight: Begin with the theorem you are allowed to use (angle at the centre is twice the angle at the circumference). Do not start by quoting the cyclic quadrilateral result, because that is what you are proving.

11. (a) Angle OAT = angle OBT = 90°, because the radius is perpendicular to the tangent [1]; angle AOB = 360° - 90° - 90° - 56° = 124°, from the angles of a quadrilateral [1] (b) Angle ACB = 124° ÷ 2 = 62°, because the angle at the centre is twice the angle at the circumference [1] [1] (c) Angle TAB = angle ACB = 62°, by the alternate segment theorem [1] [1] (d) Angle ADB = 180° - 62° = 118° [1], because opposite angles of a cyclic quadrilateral ACBD add up to 180° [1] Examiner insight: In a multi-part question, later parts often build on earlier ones. State the angle you carry forward and its reason, and check that two routes agree: here triangle TAB is isosceles with apex 56°, giving 62° as well.

Where marks are usually lost

• Giving an angle without a reason when the question asks for one.
• Writing “circle theorem” or “angles in a circle” instead of naming the theorem.
• Pairing angles that stand on different chords when using angles in the same segment.
• Using the same-side angle instead of the opposite-segment angle in the alternate segment theorem.
• Forgetting the 90° between the radius and the tangent.
• Using the cyclic quadrilateral result inside a proof of that same result.
• Proving only one direction when a statement is “if and only if”.
• Rounding part-way through a trigonometry step, or leaving off units on a length.

Next steps

• Go back to the revision notes and the study guide.
• Open the OCR GCSE Mathematics hub and tick off topics on the printable checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, GCSE (9-1) Mathematics J560, Version 2.1 (August 2026), for first assessment in 2017: section 8.05 Circles.