OxfordAQA International GCSE Mathematics: Circle properties, theorems and proof (9260) – 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 references G8 and G9 in section 3.3.1 of the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1, for exams from May/June 2018 onwards. Questions 1 and 2 use Core content. Questions 3 to 12 are Extension only, because the circle theorems and geometrical proof are listed in the Extension column. Diagrams are not drawn: sketch each one from the description. Give a reason for every angle you find.

Support: study guide, revision notes, the course hub and the checklist.

Questions

1. O is the centre of a circle. Name each of the following.

(a) A line from O to the circumference. [1] (b) A straight line that touches the circle at exactly one point. [1] (c) The longest chord. [1] (d) The region bounded by two radii and an arc. [1]

2. A and B lie on a circle with centre O. Angle OAB = 28°.

(a) Find angle AOB. [2] (b) In a different circle with centre O and diameter 15.6 cm, a chord AB has length 10.2 cm. Find the perimeter of triangle OAB. [1]

3. (Extension only) A, B and C lie on a circle with centre O, and C is on the major arc AB. Angle ACB = 67°.

(a) Find angle AOB. [1] (b) Find angle OAB. [2]

4. (Extension only) PQ is a diameter of a circle and R is on the circumference. Angle QPR = (5y − 6)° and angle PQR = (y + 12)°.

(a) State angle PRQ, with a reason. [1] (b) Find y. [2] (c) Find angle QPR. [1]

5. (Extension only) A, B, C and D lie in that order on a circle. Angle CAD = 41° and angle ABC = 107°.

(a) Find angle CBD, giving a reason. [2] (b) Find angle ADC, giving a reason. [2] (c) Find angle ACD. [2]

6. (Extension only) The tangents from T touch a circle of radius 6 cm, centre O, at A and B. OT = 10 cm.

(a) Find the length TA. [2] (b) Find angle ATO, to 1 decimal place. [2] (c) State the length TB, with a reason. [1]

7. (Extension only) A circle has centre O and radius 13 cm. Parallel chords AB and CD are on opposite sides of O. AB = 24 cm and CD = 10 cm.

(a) Find the distance from O to AB. [2] (b) Find the distance from O to CD. [2] (c) Find the distance between the chords. [1]

8. (Extension only) A tangent touches a circle at A. B and C are points on the circle on opposite sides of A. The tangent makes an angle of 58° with AB and an angle of 47° with AC. Find the three angles of triangle ABC, giving reasons. [4]

9. (Extension only) P, Q, R and S lie in that order on a circle. The tangent XPY at P, with X on the same side of P as Q, is parallel to the chord QS. Angle QPS = 76°.

(a) Show that angle PQS = 52°. Give a reason for each step. [4] (b) Find angle QRS, giving a reason. [2]

10. (Extension only) The vertices of parallelogram ABCD all lie on a circle. Prove that ABCD is a rectangle. [4]

11. (Extension only) The tangents at A and B to a circle with centre O meet at T.

(a) Prove that triangles OAT and OBT are congruent. [4] (b) Hence state two facts about TA, TB and the line OT. [1] (c) OT meets AB at M. Prove that OT is perpendicular to AB. [2]

12. (Extension only) A circle has centre O and radius 10 cm. AB is a chord of length 16 cm with midpoint M, and OM is perpendicular to AB. The tangents at A and B meet at T, and T lies on the line OM extended.

(a) Show that OM = 6 cm. [2] (b) Find angle AOM, to 1 decimal place. [2] (c) Find angle ATO, to 1 decimal place. [2] (d) Find the length AT, to 3 significant figures. [2]

Answers

1. (a) Radius [1] (b) Tangent [1] (c) Diameter [1] (d) Sector [1] Examiner insight: Give the exact word asked for. “Line” or “edge” for a tangent or chord gives no sign that you know the term, so write the precise vocabulary.

2. (a) OA = OB (radii), so angle OBA = 28° [1]; angle AOB = 180° − 28° − 28° = 124° [1] (b) Radius = 7.8 cm, so perimeter = 7.8 + 7.8 + 10.2 = 25.8 cm [1] Examiner insight: A diameter is given in part (b), so halve it before adding. Using 15.6 twice would overstate the perimeter.

3. (a) Angle AOB = 2 × 67° = 134°, angle at the centre is twice the angle at the circumference [1] (b) OA = OB (radii), so triangle OAB is isosceles [1]; angle OAB = (180° − 134°) ÷ 2 = 23° [1] Examiner insight: State which fact you use at each step. A correct angle with no reason can leave a “give a reason” part incomplete.

4. (a) Angle PRQ = 90°, angle in a semicircle [1] (b) (5y − 6) + (y + 12) + 90 = 180 [1]; 6y + 96 = 180, so y = 14 [1] (c) Angle QPR = 5 × 14 − 6 = 64° [1] Examiner insight: Build the equation from the sum of angles in the triangle, including the 90° you found in part (a). Check by testing that the two angles sum to 90°: 64° + 26°.

5. (a) Angle CBD = 41° [1], because angles in the same segment are equal (both stand on CD) [1] (b) Angle ADC = 180° − 107° = 73° [1], because opposite angles of a cyclic quadrilateral sum to 180° [1] (c) Angle ABD = 107° − 41° = 66° [1]; angle ACD = 66°, same segment on chord AD [1] Examiner insight: In multi-part angle questions, name the chord each pair of angles stands on. That makes the same-segment reason specific and easy to check.

6. (a) Angle OAT = 90°, tangent perpendicular to the radius [1]; TA² = 10² − 6² = 64, so TA = 8 cm [1] (b) sin(ATO) = 6/10 [1]; angle ATO = 36.9° [1] (c) TB = 8 cm, tangents from an external point are equal in length [1] Examiner insight: OT is the hypotenuse, so subtract the squares. Adding 6² and 10² is the usual error. Keep the unrounded value until the end when rounding to 1 d.p.

7. (a) Perpendicular from O bisects AB, so half-chord = 12 cm [1]; distance = √(13² − 12²) = 5 cm [1] (b) Half-chord = 5 cm [1]; distance = √(13² − 5²) = 12 cm [1] (c) The chords lie on opposite sides of O, so the distance is 5 + 12 = 17 cm [1] Examiner insight: Draw both chords before adding. On opposite sides the distances add. If they were on the same side, you would subtract.

8. Angle ACB = 58°, alternate segment theorem (angle between tangent and AB) [1]; angle ABC = 47°, alternate segment theorem (angle between tangent and AC) [1]; angle BAC = 180° − 58° − 47° [1] = 75° [1] Examiner insight: Match each tangent angle to the chord that forms it, then take the angle in the opposite segment. Check that angle BAC sits on the straight line between the two tangent angles, so 58° + 75° + 47° = 180°.

9. (a) Angle XPQ = angle PQS, alternate angles (XY ∥ QS) [1]; angle XPQ = angle PSQ, alternate segment theorem [1]; so angle PQS = angle PSQ, and triangle PQS is isosceles [1]; angle PQS = (180° − 76°) ÷ 2 = 52° [1] (b) Angle QRS = 180° − 76° [1] = 104°, opposite angles of cyclic quadrilateral PQRS sum to 180° [1] Examiner insight: In a “show that” the target value is given, so write every link. Quote both the parallel-line fact and the circle theorem rather than jumping to 52°.

10. Opposite angles of a parallelogram are equal, so angle A = angle C [1]; opposite angles of a cyclic quadrilateral sum to 180°, so A + C = 180° [1]; hence A = C = 90°, and in the same way B = D = 90° [1]; a parallelogram with four right angles is a rectangle [1] Examiner insight: Finish with the stated conclusion. Working that reaches 90° but never says “so ABCD is a rectangle” leaves the proof unfinished.

11. (a) OA = OB, radii [1]; angle OAT = angle OBT = 90°, tangent perpendicular to radius [1]; OT is common [1]; so the triangles are congruent (right angle, hypotenuse, side) [1] (b) TA = TB, and angle ATO = angle BTO (corresponding parts of congruent triangles) [1] (c) In triangles ATM and BTM: TA = TB, angle ATM = angle BTM, TM common, so they are congruent (two sides and the included angle) [1]; angle AMT = angle BMT, and they sum to 180° on a straight line, so each is 90° [1] Examiner insight: A congruence proof states each equal pair with its reason, then names the condition. List the three facts in the same order as the triangle letters.

12. (a) AM = 8 cm, half of 16 [1]; OM² = 10² − 8² = 36, so OM = 6 cm [1] (b) cos(AOM) = 6/10 [1]; angle AOM = 53.1° [1] (c) Angle OAT = 90° [1]; angle ATO = 180° − 90° − 53.1° = 36.9° [1] (d) tan(53.1°) = AT/10 [1]; AT = 13.3 cm [1] Examiner insight: Part (c) needs the angle at O in the larger triangle OAT, which is the same angle as in part (b). Use unrounded values for the working and round only the final answer.

Where marks are usually lost

• Leaving angles with no reason, or with a vague reason such as “circle rule”.
• Applying the angle-at-the-centre theorem to angles that stand on different arcs.
• Pairing adjacent angles of a cyclic quadrilateral instead of opposite ones.
• Reading the alternate segment on the wrong side of the chord.
• Subtracting the wrong way in right-angled triangles: the hypotenuse is OT in a tangent problem and the radius in a chord problem.
• Forgetting to double the half-chord, or halving a diameter before using it.
• Writing a proof that ends before the stated conclusion.
• Rounding in the middle of a multi-step angle or length calculation.

Next steps

• Review the revision notes and the study guide.
• Check the course hub and the printable checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class for help with proofs.

Official syllabus

These questions follow the OxfordAQA International GCSE Mathematics (9260) Specification, Version 5.1, for exams in May/June 2018 onwards, published by OxfordAQA, section 3.3.1, references G8 and G9.