IGCSE Maths 0580 Circle Theorems Past Paper Questions and Worked Solutions

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IGCSE Maths (0580) · Geometry · 4.7

Every circle theorems question from 16 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 4.7. 18 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/13/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Geometry past paper questions.

0580 Circle Theorems past paper questions: Core (Papers 1 and 3)

15 marksCore

May/June 2026 · Paper 1 · 0580/13/M/J/26 · Q9

A, B and C are points on the circumference of a circle, centre O. DAE is a tangent to the circle at A. DE is parallel to BC. BOE is a straight line. Angle AEO = 35°.

0580/13 May/June 2026 Q9 – circle centre O, tangent DAE at A, BOE a straight line, angles x° and y° marked35°x°y°OAEBCDNOT TOSCALE

(a) Write down the two geometrical reasons to explain why angle AOE = 55°.

(b) Find the value of x (angle OBC). (c) Find the value of y (angle BOC).

Show solution
Part (a)
  1. DAE is a tangent and OA is a radius, so they meet at 90°: angle OAE = 90°.
  2. The angle between a tangent and a radius is 90°
  3. Then the angles in triangle OAE add up to 180°: angle AOE = 180 − 90 − 35 = 55°.
  4. Angles in a triangle add up to 180°
Answer The angle between a tangent and a radius is 90° Angles in a triangle add up to 180°
Part (b)
  1. BC is parallel to DE, and the straight line BOE crosses both of them.
  2. Angle OBC and angle OEA are alternate angles, so they are equal: x = 35
Answer x = 35
Part (c)
  1. OB and OC are both radii, so triangle OBC is isosceles and angle OCB = angle OBC = 35°.
  2. Angles in a triangle add up to 180°: y = 180 − 35 − 35 = 110
Answer 110
23 marksCore

May/June 2026 · Paper 3 · 0580/31/M/J/26 · Q19

A, B, C and D are points on the circumference of a circle with centre O. AB = AD and angle BDC = 52°. BD is a diameter of the circle. Calculate angle ABC.

0580/31 May/June 2026 Q19 – circle centre O, diameter BD, points A and C, angle BDC = 52°52°OBDACNOT TOSCALE
Show solution
  1. BD is a diameter, so angles BAD and BCD are both angles in a semicircle: each is 90°.
  2. AB = AD, so triangle ABD is isosceles. Its base angles are equal: angle ABD = (180 − 90) ÷ 2 = 45°.
  3. In triangle BCD: angle DBC = 180 − 90 − 52 = 38° (angles in a triangle).
  4. Angle ABC = angle ABD + angle DBC = 45 + 38 = 83°
Answer 83°
31 markCore

May/June 2026 · Paper 3 · 0580/33/M/J/26 · Q25(a)

The diagram shows a circle, centre O and diameter AC. B is a point on the circumference of the circle. Give the geometrical reason why angle ABC = 90°.

0580/33 May/June 2026 Q25 – circle centre O, diameter AC, AB = 14 cm and BC = 8 cm14 cm8 cmOACBNOT TOSCALE
Show solution
  1. AC is a diameter, so triangle ABC sits in a semicircle with B on the circumference.
  2. The angle in a semicircle is 90°
Answer The angle in a semicircle is 90°
41 markCore

Oct/Nov 2025 · Paper 1 · 0580/12/O/N/25 · Q24

B is a point on the circle, centre O. ABC is a tangent to the circle at B and angle OAB = 36°. Work out the size of angle AOB.

0580/12 October/November 2025 Q24 – circle centre O with tangent ABC at B and angle 36° at A36°OBACNOT TOSCALE
Show solution
  1. A tangent meets the radius at the point of contact at 90°, so angle OBA = 90°.
  2. Angles in triangle OAB add up to 180°.
  3. Angle AOB = 180 − 90 − 36 = 54°
Answer 54°
52 marksCore

Oct/Nov 2025 · Paper 3 · 0580/31/O/N/25 · Q20(b)

A, B and C lie on the circumference of a circle, centre O. AB is a diameter. DBE is a tangent to the circle at B. Angle CBA = 71° and angle DBC = x°. Find the value of x. Give a geometrical reason for your answer.

0580/31 October/November 2025 Q20(b) – circle centre O with diameter AB and tangent DBE at B71°x°OABCDENOT TOSCALE
Show solution
  1. AB is a diameter, so OB is a radius. DBE is a tangent at B, so angle ABE (and angle ABD) = 90°.
  2. Angle DBC + angle CBA = angle DBA = 90°.
  3. x = 90 − 71, so x = 19 because the angle between a tangent and a radius is 90°
Answer x = 19 the angle between a tangent and a radius is 90°
62 marksCore

May/June 2025 · Paper 1 · 0580/11/M/J/25 · Q21

Points A, B and C lie on the circle, centre O. AC is a diameter and angle BAC = 39°. Work out angle BCA.

0580/11 May/June 2025 Q21 – circle centre O with triangle ABC and angle 39° at A39°OABCNOT TOSCALE
Show solution
  1. AC is a diameter, so ABC is a triangle in a semicircle.
  2. The angle in a semicircle is 90°, so angle ABC = 90°.
  3. Angles in a triangle add up to 180°: angle BCA = 180 − 90 − 39 = 51°
Answer 51°
75 marksCore

May/June 2025 · Paper 1 · 0580/13/M/J/25 · Q18

The diagram shows a circle, centre O. P and S are points on the circle. POR is a straight line. QRST is a tangent to the circle at S. Angle OPS = 25°, angle OSP = x° and the angle at R on the other side of the line from S is y°.

0580/13 May/June 2025 Q18 – circle centre O, tangent QRST at S, straight line POR with angle 25° at P, x° at S and y° at R25°x°y°OSPRQTNOT TOSCALE

(a) Find the value of x. Give a geometrical reason for your answer.

(b) Find the value of y.

Show solution
Part (a)
  1. OP and OS are both radii, so they are equal. Triangle OPS is isosceles.
  2. The base angles of an isosceles triangle are equal, so angle OSP = angle OPS.
  3. x = 25, because base angles of an isosceles triangle are equal
Answer 25, because base angles of an isosceles triangle are equal
Part (b)
  1. A tangent meets a radius at 90°, so angle OSR = 90°.
  2. POR is a straight line, so angle ROS is the exterior angle of triangle OPS. It equals the two opposite interior angles: 25 + 25 = 50°.
  3. Triangle ORS: angle ORS = 180 − 90 − 50 = 40°.
  4. Q, R and S are on a straight line, so y = 180 − 40 = 140
Answer 140

0580 Circle Theorems past paper questions: Extended (Papers 2 and 4)

85 marksExtended

May/June 2026 · Paper 2 · 0580/22/M/J/26 · Q15

A, B, C and D are points on a circle. EF is a tangent to the circle at A. Angle ADB = 20° and angle BCD = 75°. Find angle EAD. Give a reason for each step of your working.

0580/22 May/June 2026 Q15 – cyclic quadrilateral ABCD with tangent EF at A20°75°AEFBDCNOT TOSCALE
Show solution
  1. A, B, C and D are on the circle, so ABCD is a cyclic quadrilateral. Angles BAD and BCD are opposite.
  2. Angle BAD = 180 − 75 = 105°, because opposite angles of a cyclic quadrilateral add up to 180°
  3. Angle BAF is between the tangent AF and the chord AB. It equals the angle in the opposite segment, angle ADB.
  4. Angle BAF = 20°, by the alternate segment theorem
  5. EAF is a straight line, so the three angles at A along it add up to 180°.
  6. Angle EAD = 180 − 105 − 20 = 55°
Answer opposite angles of a cyclic quadrilateral add up to 180° alternate segment theorem 55°
91 markExtended

May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q13(b)

A, B and C are points on a circle with centre O. AOC is a straight line. Give a geometrical reason why angle ABC is 90°.

Show solution
  1. AOC is a straight line through the centre, so AC is a diameter.
  2. B is on the circle, so angle ABC is an angle in a semicircle.
  3. Reason: the angle in a semicircle is 90°
Answer the angle in a semicircle is 90°
103 marksExtended

May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q15

The diagram shows a cyclic quadrilateral ABCD. Angle ADB = 41° and angle ABC = 86°.

0580/23 May/June 2026 Q15 – cyclic quadrilateral ABCD with angles 41° and 86° marked41°86°DCBANOT TOSCALE

Find (a) angle ACB (b) angle BDC.

Show solution
Part (a)
  1. Angles ACB and ADB both stand on the same arc AB, so they are in the same segment.
  2. Angles in the same segment are equal: angle ACB = angle ADB = 41°
Answer 41°
Part (b)
  1. In triangle ABC, angle BAC = 180 − 86 − 41 = 53° (angles in a triangle).
  2. Angles BDC and BAC both stand on the arc BC, so they are equal.
  3. Angle BDC = 53°
Answer 53°
113 marksExtended

Feb/Mar 2026 · Paper 2 · 0580/22/F/M/26 · Q16(a,b)

A, B, C, D and E lie on a circle, centre O. FG is a tangent to the circle at E. Angle ABE = 35° and angle CBE = 75°. Angle AEF = x° and angle CDE = y°.

0580/22 February/March 2026 Q16 – circle centre O through A, B, C, D and E, with tangent FG at E35°75°x°y°OABCDEFGNOT TOSCALE

(a) Find the value of x and the value of y.

(b) Complete this statement with the geometrical reason: obtuse angle EOC = 150° because …

Show solution
Part (a)
  1. The alternate segment theorem says the angle between a tangent and a chord equals the angle in the opposite segment.
  2. Angle AEF (between tangent EF and chord EA) = angle ABE, so x = 35
  3. B, C, D and E all lie on the circle, so BCDE is a cyclic quadrilateral. Opposite angles CBE and CDE add up to 180°.
  4. y = 180 − 75, so y = 105
Answer x = 35 y = 105
Part (b)
  1. Angle EOC at the centre and angle CBE at the circumference both stand on the arc EC.
  2. Angle EOC = 2 × 75 = 150° because the angle at the centre is twice the angle at the circumference
Answer the angle at the centre is twice the angle at the circumference
126 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q14(a)

A, B, C, D and E lie on the circle. AC and BD intersect at X. Angle ACD = 55° and angle CXD = 88°. Complete the statements, giving a geometrical reason in each part: angle CDB = …; angle ABD = …; angle AED = …

0580/21 October/November 2025 Q14 – circle through A, B, C, D and E with chords AC and BD crossing at X55°88°CBAEDXNOT TOSCALE
Show solution
  1. Angle CDB is in triangle CXD, which has angles 55° and 88°.
  2. Angle CDB = 180 − 55 − 88 = 37° because angles in a triangle add up to 180°
  3. Angles ABD and ACD both stand on the same arc AD.
  4. Angle ABD = angle ACD = 55° because angles in the same segment are equal
  5. ACDE is a quadrilateral with all four corners on the circle, so it is cyclic. Angle AED is opposite angle ACD.
  6. Angle AED = 180 − 55 = 125° because opposite angles of a cyclic quadrilateral (ACDE) add up to 180°
Answer 37° angles in a triangle add up to 180° 55° angles in the same segment are equal 125° opposite angles of a cyclic quadrilateral (ACDE) add up to 180°
134 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q13

A, B, C and D are points on the circumference of a circle with centre O. ED and EB are tangents to the circle. AC is a diameter and AC is parallel to EB. Angle AOD = 44° and angle OEB = x°.

0580/23 October/November 2025 Q13 – circle centre O with tangents ED and EB, diameter AC parallel to EB, angle AOD = 44°44°x°OACDBENOT TOSCALE

Find the value of x.

Show solution
  1. ED and EB are tangents from the same point E. OE cuts the angle between them in half, so angle OED = angle OEB = x°.
  2. AC is parallel to EB, and OE crosses both. So angle AOE = angle OEB = x° (alternate angles).
  3. Angle DOE = angle DOA + angle AOE = 44 + x.
  4. A tangent meets a radius at 90°, so angle ODE = 90°.
  5. Angles in triangle ODE add up to 180°: x + (44 + x) + 90 = 180.
  6. Simplify: 2x + 134 = 180, so 2x = 46 and x = 23
Answer x = 23
144 marksExtended

Oct/Nov 2025 · Paper 2 · 0580/23/O/N/25 · Q21

The diagram shows a cyclic quadrilateral with angles y°, m°, p° and (m + 10)°, where p and m are opposite angles and y and (m + 10) are opposite angles. The ratio p : m = 2 : 3. Find the value of y.

0580/23 October/November 2025 Q21 – cyclic quadrilateral with angles y°, m°, p° and (m + 10)°y°m°(m + 10)°p°NOT TOSCALE
Show solution
  1. Opposite angles of a cyclic quadrilateral add up to 180°, so p + m = 180.
  2. p : m = 2 : 3 has 5 parts. One part = 180 ÷ 5 = 36, so m = 3 × 36 = 108.
  3. The angle (m + 10)° = 108 + 10 = 118°.
  4. y is opposite (m + 10)°, so y = 180 − 118 = 62
Answer 62
153 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q15

A, B and C lie on a circle, centre O. TA and TB are tangents to the circle at A and B. Angle ACB = 67°. Calculate angle ATB.

0580/41 October/November 2025 Q15 – circle centre O with tangents TA and TB, angle 67° at C67°OBACTNOT TOSCALE
Show solution
  1. The angle at the centre is twice the angle at the circumference, so angle AOB = 2 × 67 = 134°.
  2. A tangent meets a radius at 90°, so angle OAT = 90° and angle OBT = 90°.
  3. OATB is a quadrilateral, and the angles in a quadrilateral add up to 360°.
  4. Angle ATB = 360 − 90 − 90 − 134 = 46°
Answer 46°
164 marksExtended

Oct/Nov 2025 · Paper 4 · 0580/42/O/N/25 · Q12

A, B and C lie on a circle centre O. DE is a tangent to the circle at C. Angle ABC = 52° and angle BCE = 65°. Angle AOC = x° and angle OAB = y°.

0580/42 October/November 2025 Q12 – circle centre O with tangent DE at C, angles 52° and 65°52°65°x°y°OCBADENOT TOSCALE

(a) Find the value of x. Give a geometrical reason. (b) Find the value of y.

Show solution
Part (a)
  1. Angle AOC is at the centre and angle ABC is at the circumference, both standing on the arc AC.
  2. x = 2 × 52 = 104, so x = 104 because the angle at the centre is twice the angle at the circumference
Answer x = 104 the angle at the centre is twice the angle at the circumference
Part (b)
  1. Alternate segment theorem: the angle between the tangent and the chord CB equals the angle in the other segment. So angle BAC = angle BCE = 65°.
  2. OA = OC (both radii), so triangle OAC is isosceles.
  3. Angle OAC = (180 − 104) ÷ 2 = 76 ÷ 2 = 38°.
  4. Angle OAB = angle BAC − angle OAC = 65 − 38 = 27
Answer 27
174 marksExtended

May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q10

A, B and C are three points on a circle, centre O. DE is a tangent to the circle at A. Angle ACO = 35° and angle BCO = 40°.

0580/21 May/June 2025 Q10 – circle centre O with tangent DE at A, points B and C, angles 35° and 40° at C35°40°OADECBNOT TOSCALE

Find (a) angle AOC (b) angle ABC (c) angle DAC (d) angle OAB.

Show solution
Part (a)
  1. OA and OC are radii, so triangle OAC is isosceles and angle OAC = angle OCA = 35°.
  2. Angles in a triangle add up to 180°: angle AOC = 180 − 35 − 35 = 110°
Answer 110°
Part (b)
  1. The angle at the centre is twice the angle at the circumference, both standing on arc AC.
  2. Angle ABC = 110 ÷ 2 = 55°
Answer 55°
Part (c)
  1. By the alternate segment theorem, the angle between the tangent DA and the chord AC equals the angle in the opposite segment, angle ABC.
  2. Angle DAC = 55°
Answer 55°
Part (d)
  1. OB and OC are radii, so triangle OBC is isosceles and angle OBC = angle OCB = 40°.
  2. Angle OBA = angle ABC − angle OBC = 55 − 40 = 15°.
  3. OA and OB are radii, so triangle OAB is isosceles and angle OAB = angle OBA.
  4. Angle OAB = 15°
Answer 15°
182 marksExtended

May/June 2025 · Paper 2 · 0580/22/M/J/25 · Q12

P, Q and R lie on a circle. QR is a diameter. Angle PQR = 74°. Find angle PRQ. Give geometrical reasons for your answer.

0580/22 May/June 2025 Q12 – circle with diameter QR, point P and angle 74° at Q74°RQPNOT TOSCALE
Show solution
  1. QR is a diameter, so the angle QPR is in a semicircle: angle RPQ = 90° (the angle in a semicircle is 90°).
  2. Angles in triangle PQR add up to 180°: angle PRQ = 180 − 90 − 74 = 16°.
  3. 16°, because the angle in a semicircle is 90° and angles in a triangle add up to 180°
Answer 16°, because the angle in a semicircle is 90° and angles in a triangle add up to 180°

Question papers used on this page

All circle theorems questions above, listed by 0580 question paper code.

QP codeFile namePaperQuestions
0580/13/M/J/260580_s26_qp_13May/June 2026 · Paper 1 (Core)Q9
0580/22/M/J/260580_s26_qp_22May/June 2026 · Paper 2 (Extended)Q15
0580/23/M/J/260580_s26_qp_23May/June 2026 · Paper 2 (Extended)Q13(b), Q15
0580/31/M/J/260580_s26_qp_31May/June 2026 · Paper 3 (Core)Q19
0580/33/M/J/260580_s26_qp_33May/June 2026 · Paper 3 (Core)Q25(a)
0580/22/F/M/260580_m26_qp_22Feb/Mar 2026 · Paper 2 (Extended)Q16(a,b)
0580/12/O/N/250580_w25_qp_12Oct/Nov 2025 · Paper 1 (Core)Q24
0580/21/O/N/250580_w25_qp_21Oct/Nov 2025 · Paper 2 (Extended)Q14(a)
0580/23/O/N/250580_w25_qp_23Oct/Nov 2025 · Paper 2 (Extended)Q13, Q21
0580/31/O/N/250580_w25_qp_31Oct/Nov 2025 · Paper 3 (Core)Q20(b)
0580/41/O/N/250580_w25_qp_41Oct/Nov 2025 · Paper 4 (Extended)Q15
0580/42/O/N/250580_w25_qp_42Oct/Nov 2025 · Paper 4 (Extended)Q12
0580/11/M/J/250580_s25_qp_11May/June 2025 · Paper 1 (Core)Q21
0580/13/M/J/250580_s25_qp_13May/June 2025 · Paper 1 (Core)Q18
0580/21/M/J/250580_s25_qp_21May/June 2025 · Paper 2 (Extended)Q10
0580/22/M/J/250580_s25_qp_22May/June 2025 · Paper 2 (Extended)Q12

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