Edexcel International GCSE Maths Circle Theorems Past Paper Questions and Solutions (4MA1 Higher)

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Edexcel International GCSE Maths A (4MA1) · Higher · Geometry and Trigonometry

Circle Theorems: Edexcel 4MA1 Past Paper Questions and Worked Solutions

This page collects every circle theorems question from the Pearson Edexcel International GCSE Mathematics A (4MA1) Higher tier Paper 1H and Paper 1HR, January 2020 to June 2025: 9 questions worth 37 marks. Questions run from easier to harder. Try each one first, then click “Show solution” for a step-by-step method with the reason for every step. All answers are checked against the official Pearson mark schemes.

15 marksHigherJune 2024 · Paper 1HR (4MA1/1HR) · Q13

A, B, C and D are points on a circle, centre O. Angle BCD = 128°. Work out the size of angle OBD. Give a reason for each stage of your working.

ABCDO128°
Show solution
  1. ABCD is a cyclic quadrilateral, so opposite angles add to 180°: angle BAD = 180° − 128° = 52° (opposite angles of a cyclic quadrilateral add up to 180°).
  2. Angle BOD = 2 × 52° = 104° (the angle at the centre is twice the angle at the circumference).
  3. OB = OD because both are radii, so triangle OBD is isosceles (base angles of an isosceles triangle are equal).
  4. Angle OBD = (180° − 104°) ÷ 2 = 76° ÷ 2 (angles in a triangle add up to 180°)

Answer: 38°

24 marksHigherJanuary 2020 · Paper 1HR (4MA1/1HR) · Q14

A, B and C are points on a circle, centre O. Angle ABC = 38°. Work out the size of angle OAC. Give a reason for each stage of your working.

ABCO38°
Show solution
  1. Angle AOC = 2 × 38° = 76° (the angle at the centre is twice the angle at the circumference).
  2. OA = OC (radii), so triangle OAC is isosceles (base angles of an isosceles triangle are equal).
  3. Angle OAC = (180° − 76°) ÷ 2 (angles in a triangle add up to 180°)

Answer: 52°

34 marksHigherJanuary 2021 · Paper 1HR (4MA1/1HR) · Q14

B, D, E and F are points on a circle. ABC is the tangent at B to the circle. Angle ABD = 39° and angle EFD = 18°. Work out the size of angle BDE. Give reasons for your working.

EFDBAC18°39°
Show solution
  1. Angle BFD = angle ABD = 39° (alternate segment theorem: the angle between the tangent and chord BD equals the angle in the alternate segment).
  2. Angle BFE = angle BFD + angle DFE = 39° + 18° = 57°
  3. BDEF is a cyclic quadrilateral, so angle BDE = 180° − 57° (opposite angles of a cyclic quadrilateral add up to 180°)

Answer: 123°

44 marksHigherJanuary 2022 · Paper 1HR (4MA1/1HR) · Q14

A, B, C and D are points on a circle, centre O. AOC is a diameter of the circle. Angle BAC = 55°. Work out the size of angle ADB. Give a reason for each stage of your working.

ABCDO55°
Show solution
  1. Angle ABC = 90° (the angle in a semicircle is 90°, because AC is a diameter).
  2. Angle ACB = 180° − 90° − 55° = 35° (angles in a triangle add up to 180°).
  3. Angle ADB = angle ACB (angles in the same segment are equal: both stand on the chord AB).

Answer: 35°

54 marksHigherJune 2020 · Paper 1HR (4MA1/1HR) · Q15

B, D, E and F are points on a circle. ABC is the tangent to the circle at B. Angle EDF = 40° and angle FBC = 70°. Prove that the tangent ABC is parallel to EF. Give a reason for each stage of your working.

EFDBAC40°70°
Show solution
  1. Angle EBF = angle EDF = 40° (angles in the same segment are equal: both are subtended by the chord EF). [Join BE to see this angle.]
  2. Angle BEF = angle FBC = 70° (alternate segment theorem).
  3. In triangle BEF: angle BFE = 180° − 40° − 70° = 70° (angles in a triangle add up to 180°).
  4. So angle BFE = angle FBC = 70°. These are alternate angles between the lines EF and ABC, with BF crossing them.

Answer: Alternate angles BFE and FBC are both 70°, so ABC is parallel to EF

64 marksHigherJune 2021 · Paper 1H (4MA1/1H) · Q15

P, Q and R are points on a circle, centre O. TRV is the tangent to the circle at R. Reflex angle POR = 238°. Angle QRV = 60°. Calculate the size of angle OPQ. Give a reason for each stage of your working.

QPRVTO238°60°
Show solution
  1. Angle POR = 360° − 238° = 122° (angles around a point add up to 360°).
  2. OP = OR (radii), so triangle OPR is isosceles: angle OPR = (180° − 122°) ÷ 2 = 29° (base angles of an isosceles triangle are equal; angles in a triangle add to 180°).
  3. Angle QPR = angle QRV = 60° (alternate segment theorem: the angle between a tangent and a chord equals the angle in the alternate segment).
  4. Angle OPQ = angle QPR − angle OPR = 60° − 29°

Answer: 31°

73 marksHigherJune 2022 · Paper 1HR (4MA1/1HR) · Q15

P, Q, R and S are points on a circle with centre O. PS is a diameter of the circle. Angle PQR = 136°. Work out the size of angle RPS.

PQRSO136°
Show solution
  1. PQRS is a cyclic quadrilateral, so opposite angles add up to 180°: angle PSR = 180° − 136° = 44°
  2. PS is a diameter, so angle PRS = 90° (the angle in a semicircle is 90°).
  3. Angles in triangle PRS add up to 180°: angle RPS = 180° − 90° − 44°

Answer: 46°

84 marksHigherJune 2020 · Paper 1H (4MA1/1H) · Q16

L, M, N and P are points on a circle, centre O. Angle MNP = 58°.

NPLMO58°

(a)(i) Find the size of angle MLP. (ii) Give a reason for your answer. (2)

(b) Find the size of the reflex angle MOP. (2)

Show solution

Part (a)

  1. LMNP is a cyclic quadrilateral and angle MLP is opposite angle MNP.
  2. 180° − 58° = 122°

Answer: 122°, because opposite angles of a cyclic quadrilateral add up to 180°

Part (b)

  1. The angle at the centre is twice the angle at the circumference: angle MOP = 2 × 58° = 116°
  2. Reflex angle MOP = 360° − 116°

Answer: 244°

95 marksHigherJune 2025 · Paper 1HR (4MA1/1HR) · Q23

PTR and QTS are chords of a circle that cross at T. PT = (3y − 5) cm, QT = (y + 3) cm, RT = 4x cm and ST = (x + 2) cm. Find an expression for y in terms of x.

PRQST(3y − 5)4x(y + 3)(x + 2)
Show solution
  1. Intersecting chords theorem: when two chords cross, PT × TR = QT × TS.
  2. (3y − 5) × 4x = (y + 3)(x + 2)
  3. Expand both sides: 12xy − 20x = xy + 2y + 3x + 6
  4. Collect the y terms on one side: 12xy − xy − 2y = 3x + 6 + 20x, so 11xy − 2y = 23x + 6
  5. Factorise out y: y(11x − 2) = 23x + 6

Answer: y = 23x + 6⁄11x − 2

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