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study guides18 min read

Circle Theorems: GCSE & IGCSE Practice Questions with Worked Solutions

Sir Faraz Hassan

Sir Faraz Hassan

10 Jul 2026

Table of Contents▾
    Grade 9 TargetedFourteen exam-style questions with full worked solutions and mark schemes, covering every circle theorem on the GCSE and IGCSE specifications, from the angle in a semicircle to the alternate segment theorem

    Circle theorems reward students who can spot the right rule and chain a few steps together. These fourteen questions build from single-theorem angle chases up to full multi-theorem proofs. Each one has a complete worked solution and mark scheme, so you can see exactly how every mark is earned.

    What are circle theorems?

    Circle theorems are a set of rules that link the angles, chords, tangents and radii of a circle. The eight main circle theorems are: the angle in a semicircle is 90 degrees; the angle at the centre is twice the angle at the circumference; angles in the same segment are equal; opposite angles of a cyclic quadrilateral add up to 180 degrees; a tangent meets a radius at 90 degrees; the two tangents from a point are equal in length; the perpendicular from the centre bisects a chord; and the alternate segment theorem, which says the angle between a tangent and a chord equals the angle in the alternate segment. They appear on every major GCSE and IGCSE specification: on the UK GCSE Higher tier for AQA, Edexcel and OCR, on Cambridge IGCSE 0580 as spec points C4.7, E4.7 and E4.8, and on Edexcel International GCSE 4MA1 as section 4.6. Most exam questions combine two or more of these, so the real skill is recognising which theorem applies from the diagram and chaining the steps together.

    Which exam boards is this on?

    Circle theorems are on every major GCSE and IGCSE specification, but how much you need depends on your board and your tier. The lower tier is not empty. Both IGCSE boards put a small number of circle properties on the Core or Foundation paper, and save the angle theorems for the top tier.

    UK GCSE: AQA, Edexcel and OCR

    Circle theorems are Higher tier only. They do not appear on a Foundation paper. The content is set by the Department for Education and is the same across all three boards, so only the question style differs. They are usually worth 3 to 5 marks and sit from Grade 7 up to Grade 9, with the alternate segment theorem and multi-step proofs at the top end.

    Cambridge IGCSE 0580: spec points C4.7, E4.7 and E4.8

    Cambridge splits circle theorems across both tiers. Core has C4.7, which is limited to just two: the angle in a semicircle is 90 degrees, and the angle between a tangent and a radius is 90 degrees. C4.8 is Extended content only. Extended has E4.7, Circle theorems I, which adds the angle at the centre, angles in the same segment, cyclic quadrilaterals and the alternate segment theorem, and E4.8, Circle theorems II, which covers the three symmetry properties: equal chords are equidistant from the centre, the perpendicular bisector of a chord passes through the centre, and tangents from an external point are equal in length. Cambridge expects you to name the property when you give a reason.

    Edexcel International GCSE 4MA1: section 4.6

    Edexcel also splits them. Foundation section 4.6 covers three chord and tangent properties: two tangents from a point are equal in length, a tangent is perpendicular to the radius at the point of contact, and the line from the centre perpendicular to a chord bisects that chord. Higher section 4.6 adds the angle properties, the cyclic quadrilateral, the alternate segment theorem, and one thing no other board asks for: the internal and external intersecting chord properties. Formal proof of the theorems is not required on 4MA1, but reasons are.

    Calculator, or not?

    This differs by board, and it changes how you should practise. On Cambridge IGCSE 0580, Paper 1 (Core) and Paper 2 (Extended) are non-calculator, and Papers 3 and 4 allow one. Both non-calculator papers are new: Cambridge introduced them for the 2025 exams, and before that every 0580 paper allowed a calculator. On Edexcel International GCSE 4MA1 there is no non-calculator paper at all, so a calculator is allowed in both. On the UK GCSE, Paper 1 is non-calculator and Papers 2 and 3 allow one. It is worth checking which papers you sit before you decide how much of this to practise without a calculator.

    A note on grades

    Cambridge IGCSE 0580 is graded A* to G, not 9 to 1. The Extended tier is eligible for grades A* to E, and the Core tier for C to G. The grade badges on this page use the 9 to 1 scale, which is what the UK GCSE and Edexcel International GCSE 4MA1 use.

    How to use this page

    1

    Try each question on paper first

    Give yourself a real attempt before looking at the solution, because that is where the learning happens.

    2

    Check the calculator icon

    A crossed-out calculator means non-calculator; a plain calculator means a calculator is allowed.

    3

    Reveal the worked solution

    Open the solution under each question and compare it with your own method, step by step.

    4

    Read the mark scheme

    See exactly where each method mark (M1) and accuracy mark (A1) is awarded, and remember that at GCSE the theorem reasons carry marks too.

    5

    Come back and redo it

    Repeat any question you got wrong a few days later to lock the method in.

    Download printable PDF

    14 questions with full worked solutions and mark schemes - free PDF

    Practice questions

    Work through each question, then open the worked solution to check your method. Single-theorem questions come first, then multi-theorem problems and proofs.

    Question 1, Grade 8, Non-calculator

    Circle centre O with diameter PR; points Q and S on the circumference, angle SPR = 54 degreesPQRSO54°xNot drawn accurately

    PP, QQ, RR and SS are points on a circle, centre OO. PORPOR is a straight line, so PRPR is a diameter. ∠SPR=54∘\angle SPR = 54^{\circ}. Work out the size of the angle marked xx (∠PQS\angle PQS).

    x=x =∘^{\circ}
    [Total 3 marks]
    GRADE8
    Show solution & mark schemeHide solution & mark scheme

    Question 1 - Exam Solution

    Understanding the Question
    Given
    P,Q,R,SP, Q, R, S lie on a circle, centre OO; PORPOR is a straight line, so PRPR is a diameter.
    ∠SPR=54∘\angle SPR = 54^{\circ}.
    Find
    The size of the angle marked xx (∠PQS\angle PQS).
    Plan the Solution
    • PRPR is a diameter, so ∠PQR\angle PQR is the angle in a semicircle and equals 90∘90^{\circ}.
    • Use angles in the same segment to find ∠SQR\angle SQR.
    • Subtract to find xx.
    Worked Solution [3 marks]
    Rule - Angle chasing: apply the angle in a semicircle and angles in the same segment, then subtract to reach the required angle.
    Step 1: Use the diameter (angle in a semicircle).
    ∠PQR=90∘\angle PQR = 90^{\circ}
    (Reason: PRPR is a diameter, and the angle in a semicircle is 90∘90^{\circ}.)
    Step 2: Angles in the same segment.
    ∠SQR=∠SPR=54∘\angle SQR = \angle SPR = 54^{\circ}
    (Reason: ∠SQR\angle SQR and ∠SPR\angle SPR both stand on the chord SRSR, so angles in the same segment are equal.)
    Step 3: Subtract to find x.
    x=∠PQR−∠SQR=90∘−54∘=36∘x = \angle PQR - \angle SQR = 90^{\circ} - 54^{\circ} = 36^{\circ}
    (Reason: xx is the part of ∠PQR\angle PQR that remains after ∠SQR\angle SQR is removed.)
    x=36∘x = 36^{\circ}
    Verification
    Check 1: ∠PQS+∠SQR=36∘+54∘=90∘=∠PQR\angle PQS + \angle SQR = 36^{\circ} + 54^{\circ} = 90^{\circ} = \angle PQR
    Check 2: x=36∘x = 36^{\circ} is acute and smaller than the right angle at QQ, which fits the diagram.
    Mark Scheme Breakdown
    StepMarkDescriptionGot it?
    ∠SQR=54∘\angle SQR = 54^{\circ}M1Method - angles in the same segment✓
    ∠PQR=90∘\angle PQR = 90^{\circ}M1Method - the angle in a semicircle✓
    x=36∘x = 36^{\circ}A1Accuracy - correct final answer✓

    Full marks: 3/3

    Question 2, Grade 8, Non-calculator

    Circle centre O with external point T; TP and TS are tangents, PQR is a straight line, angle STO = 34 degreesPSQTRO34°Not drawn accurately

    The diagram shows a circle, centre OO. PP, QQ and SS are points on the circle. TPTP and TSTS are tangents to the circle, and PQRPQR is a straight line. ∠STO=34∘\angle STO = 34^{\circ}. Work out the size of ∠SQR\angle SQR, giving reasons.

    ∠SQR=\angle SQR =∘^{\circ}
    [Total 5 marks]
    GRADE8Problem solving
    Show solution & mark schemeHide solution & mark scheme

    Question 2 - Exam Solution

    Understanding the Question
    Given
    Circle, centre OO; P,Q,SP, Q, S on the circle.
    TPTP and TSTS are tangents from TT; PQRPQR is a straight line. ∠STO=34∘\angle STO = 34^{\circ}.
    Find
    The size of ∠SQR\angle SQR.
    Plan the Solution
    • Tangent meets radius: ∠TSO=90∘\angle TSO = 90^{\circ}, then find ∠TOS\angle TOS in triangle TSOTSO.
    • The two equal tangents double the angle at the centre: ∠POS=2×∠TOS\angle POS = 2 \times \angle TOS.
    • Angle at the centre is twice the angle at the circumference: find ∠PQS\angle PQS.
    • Angles on a straight line: find ∠SQR\angle SQR.
    Worked Solution [5 marks]
    Rule - Work in from the tangent: use the tangent-radius right angle, the symmetry of the two equal tangents, and the centre-to-circumference link, then finish on the straight line.
    Step 1: Tangent meets radius.
    ∠TSO=90∘\angle TSO = 90^{\circ}
    (Reason: OSOS is a radius and TSTS is a tangent; a tangent meets a radius at 90∘90^{\circ}.)
    Step 2: Angles in triangle TSO.
    ∠TOS=180∘−90∘−34∘=56∘\angle TOS = 180^{\circ} - 90^{\circ} - 34^{\circ} = 56^{\circ}
    (Reason: The angles in triangle TSOTSO add up to 180∘180^{\circ}.)
    Step 3: Two equal tangents (angle at the centre).
    ∠POS=2×56∘=112∘\angle POS = 2 \times 56^{\circ} = 112^{\circ}
    (Reason: TPTP and TSTS are equal tangents from TT, so triangles TPOTPO and TSOTSO are congruent; this doubles the angle at OO.)
    Step 4: Angle at the centre is twice the angle at the circumference.
    ∠PQS=12×112∘=56∘\angle PQS = \tfrac{1}{2} \times 112^{\circ} = 56^{\circ}
    (Reason: ∠POS\angle POS is at the centre and ∠PQS\angle PQS is at the circumference, both standing on arc PSPS.)
    Step 5: Angles on a straight line.
    ∠SQR=180∘−56∘=124∘\angle SQR = 180^{\circ} - 56^{\circ} = 124^{\circ}
    (Reason: PQRPQR is a straight line, so the angles at QQ add up to 180∘180^{\circ}.)
    ∠SQR=124∘\angle SQR = 124^{\circ}
    Verification
    Check 1: In triangle TSOTSO: 90∘+34∘+56∘=180∘90^{\circ} + 34^{\circ} + 56^{\circ} = 180^{\circ}.
    Check 2: Quadrilateral OPTSOPTS: 90∘+68∘+90∘+112∘=360∘90^{\circ} + 68^{\circ} + 90^{\circ} + 112^{\circ} = 360^{\circ} (with ∠PTS=2×34∘=68∘\angle PTS = 2 \times 34^{\circ} = 68^{\circ}).
    Check 3: At QQ: ∠PQS+∠SQR=56∘+124∘=180∘\angle PQS + \angle SQR = 56^{\circ} + 124^{\circ} = 180^{\circ}, matching the straight line.
    Mark Scheme Breakdown
    StepMarkDescriptionGot it?
    ∠TSO=90∘\angle TSO = 90^{\circ}M1Method - tangent meets a radius at 90 degrees✓
    ∠TOS=56∘\angle TOS = 56^{\circ}M1Method - angles in a triangle✓
    ∠POS=112∘\angle POS = 112^{\circ}M1Method - equal tangents give the angle at the centre✓
    ∠PQS=56∘\angle PQS = 56^{\circ}M1Method - angle at the centre is twice the circumference✓
    ∠SQR=124∘\angle SQR = 124^{\circ}A1Accuracy - correct final answer✓

    Full marks: 5/5

    Question 3, Grade 9, Non-calculator

    Circle with points P, Q, R, S; HK tangent at Q and NPQ a straight line; angle HQP = 108 and angle NPR = 152 degreesPQRSNHK108°152°Not drawn accurately

    PP, QQ, RR and SS lie on a circle. NPQNPQ is a straight line and HKHK is the tangent to the circle at QQ. ∠HQP=108∘\angle HQP = 108^{\circ} and ∠NPR=152∘\angle NPR = 152^{\circ}. Find the size of ∠RSP\angle RSP, giving a reason for each step.

    ∠RSP=\angle RSP =∘^{\circ}
    [Total 4 marks]
    GRADE9
    Show solution & mark schemeHide solution & mark scheme

    Question 3 - Exam Solution

    Understanding the Question
    Given
    P,Q,R,SP, Q, R, S lie on a circle. NPQNPQ is a straight line and HKHK is the tangent at QQ.
    ∠HQP=108∘\angle HQP = 108^{\circ} and ∠NPR=152∘\angle NPR = 152^{\circ}.
    Find
    The size of ∠RSP\angle RSP, giving a reason for each step.
    Plan the Solution
    • Alternate segment theorem: the tangent-chord angle ∠HQP\angle HQP equals ∠QRP\angle QRP.
    • Angles on a straight line at PP: find ∠RPQ\angle RPQ.
    • Angles in triangle PQRPQR: find ∠PQR\angle PQR.
    • Angles in the same segment: ∠RSP=∠PQR\angle RSP = \angle PQR.
    Worked Solution [4 marks]
    Rule - Transfer angles round the circle: use the alternate segment theorem, a straight line and a triangle, then move the angle to the required point with the same-segment rule.
    Step 1: Alternate segment theorem.
    ∠QRP=108∘\angle QRP = 108^{\circ}
    (Reason: ∠HQP\angle HQP is the angle between the tangent HKHK and the chord QPQP; by the alternate segment theorem it equals ∠QRP\angle QRP in the alternate segment.)
    Step 2: Angles on a straight line at P.
    ∠RPQ=180∘−152∘=28∘\angle RPQ = 180^{\circ} - 152^{\circ} = 28^{\circ}
    (Reason: NPQNPQ is a straight line, so ∠NPR\angle NPR and ∠RPQ\angle RPQ add up to 180∘180^{\circ}.)
    Step 3: Angles in triangle PQR.
    ∠PQR=180∘−108∘−28∘=44∘\angle PQR = 180^{\circ} - 108^{\circ} - 28^{\circ} = 44^{\circ}
    (Reason: The angles in triangle PQRPQR add up to 180∘180^{\circ}.)
    Step 4: Angles in the same segment.
    ∠RSP=∠PQR=44∘\angle RSP = \angle PQR = 44^{\circ}
    (Reason: ∠RSP\angle RSP and ∠PQR\angle PQR both stand on the chord RPRP, so angles in the same segment are equal.)
    ∠RSP=44∘\angle RSP = 44^{\circ}
    Verification
    Check 1: Triangle PQRPQR: 108∘+28∘+44∘=180∘108^{\circ} + 28^{\circ} + 44^{\circ} = 180^{\circ}.
    Check 2: Straight line NPQNPQ at PP: ∠NPR+∠RPQ=152∘+28∘=180∘\angle NPR + \angle RPQ = 152^{\circ} + 28^{\circ} = 180^{\circ}.
    Check 3: The same-segment angles agree: ∠RSP=∠PQR=44∘\angle RSP = \angle PQR = 44^{\circ}.
    Mark Scheme Breakdown
    StepMarkDescriptionGot it?
    ∠QRP=108∘\angle QRP = 108^{\circ}M1Method - alternate segment theorem✓
    ∠RPQ=28∘\angle RPQ = 28^{\circ}M1Method - angles on a straight line✓
    ∠PQR=44∘\angle PQR = 44^{\circ}M1Method - angles in a triangle✓
    ∠RSP=44∘\angle RSP = 44^{\circ}A1Accuracy - correct final answer✓

    Full marks: 4/4

    Question 4, Grade 8, Non-calculator

    Circle centre O with external point T; PT and RT tangents, P Q R S on the circle; angle PTR = 72 degreesPQRSTO72°Not drawn accurately

    The diagram shows a circle, centre OO. PP, QQ, RR and SS are points on the circle. PTPT and RTRT are tangents from TT. ∠PTR=72∘\angle PTR = 72^{\circ}. Find the size of ∠PQR\angle PQR, giving reasons.

    ∠PQR=\angle PQR =∘^{\circ}
    [Total 4 marks]
    GRADE8Problem solving
    Show solution & mark schemeHide solution & mark scheme

    Question 4 - Exam Solution

    Understanding the Question
    Given
    Circle, centre OO; P,Q,R,SP, Q, R, S on the circle.
    PTPT and RTRT are tangents from TT. ∠PTR=72∘\angle PTR = 72^{\circ}.
    Find
    The size of ∠PQR\angle PQR, giving reasons.
    Plan the Solution
    • Tangent meets radius: both ∠OPT\angle OPT and ∠ORT\angle ORT are 90∘90^{\circ}.
    • Angles in quadrilateral OPTROPTR add to 360∘360^{\circ}: find the central angle ∠POR\angle POR.
    • Angle at the centre is twice the angle at the circumference: find ∠PSR\angle PSR.
    • Opposite angles of a cyclic quadrilateral: find ∠PQR\angle PQR.
    Worked Solution [4 marks]
    Rule - From the tangents to the far angle: use the two tangent-radius right angles inside the quadrilateral, then the centre-to-circumference link, and finish with the cyclic-quadrilateral rule.
    Step 1: Tangent meets radius (two right angles).
    ∠OPT=∠ORT=90∘\angle OPT = \angle ORT = 90^{\circ}
    (Reason: OPOP and OROR are radii and PTPT, RTRT are tangents; a tangent meets a radius at 90∘90^{\circ}.)
    Step 2: Angles in quadrilateral OPTR.
    ∠POR=360∘−90∘−90∘−72∘=108∘\angle POR = 360^{\circ} - 90^{\circ} - 90^{\circ} - 72^{\circ} = 108^{\circ}
    (Reason: OPTROPTR is a quadrilateral, and its angles add up to 360∘360^{\circ}.)
    Step 3: Angle at the centre is twice the angle at the circumference.
    ∠PSR=12×108∘=54∘\angle PSR = \tfrac{1}{2} \times 108^{\circ} = 54^{\circ}
    (Reason: ∠POR\angle POR is at the centre and ∠PSR\angle PSR is at the circumference, both standing on arc PRPR.)
    Step 4: Opposite angles of a cyclic quadrilateral.
    ∠PQR=180∘−54∘=126∘\angle PQR = 180^{\circ} - 54^{\circ} = 126^{\circ}
    (Reason: PQRSPQRS is a cyclic quadrilateral, so the opposite angles ∠PQR\angle PQR and ∠PSR\angle PSR add up to 180∘180^{\circ}.)
    ∠PQR=126∘\angle PQR = 126^{\circ}
    Verification
    Check 1: Quadrilateral OPTROPTR: 90∘+90∘+72∘+108∘=360∘90^{\circ} + 90^{\circ} + 72^{\circ} + 108^{\circ} = 360^{\circ}.
    Check 2: Opposite angles: ∠PQR+∠PSR=126∘+54∘=180∘\angle PQR + \angle PSR = 126^{\circ} + 54^{\circ} = 180^{\circ}.
    Check 3: The angle at the centre (108∘108^{\circ}) is exactly twice the angle at the circumference (54∘54^{\circ}).
    Mark Scheme Breakdown
    StepMarkDescriptionGot it?
    ∠OPT=∠ORT=90∘\angle OPT = \angle ORT = 90^{\circ}M1Method - tangent meets a radius at 90 degrees✓
    ∠POR=108∘\angle POR = 108^{\circ}M1Method - angles in a quadrilateral add to 360 degrees✓
    ∠PSR=54∘\angle PSR = 54^{\circ}M1Method - angle at the centre is twice the circumference✓
    ∠PQR=126∘\angle PQR = 126^{\circ}A1Accuracy - correct final answer✓

    Full marks: 4/4

    Question 5, Grade 9, Non-calculator

    Circle with points P, Q, R, S, T; MN tangent at T; angle PQT = 34 and angle SRT = 56 degrees; chord PS is a diameterTPSQRMN34°56°Not drawn accurately

    PP, QQ, RR, SS and TT lie on a circle. MNMN is the tangent to the circle at TT. ∠PQT=34∘\angle PQT = 34^{\circ} and ∠SRT=56∘\angle SRT = 56^{\circ}. Prove that the chord PSPS passes through the centre of the circle.

    [Total 3 marks]
    GRADE9Problem solving
    Show solution & mark schemeHide solution & mark scheme

    Question 5 - Exam Solution

    Understanding the Question
    Given
    P,Q,R,S,TP, Q, R, S, T lie on a circle. MNMN is the tangent at TT.
    ∠PQT=34∘\angle PQT = 34^{\circ} and ∠SRT=56∘\angle SRT = 56^{\circ}.
    Prove
    That the chord PSPS passes through the centre of the circle.
    Plan the Solution
    • Alternate segment theorem: find the two tangent-chord angles ∠PTM\angle PTM and ∠STN\angle STN.
    • Angles on the straight line MTNMTN: find ∠PTS\angle PTS.
    • If ∠PTS=90∘\angle PTS = 90^{\circ}, then PSPS is a diameter (angle in a semicircle), so it passes through the centre.
    Worked Solution [3 marks]
    Rule - A right angle means a diameter: show that the angle at TT is 90∘90^{\circ}; the converse of the angle in a semicircle then forces PSPS to be a diameter through the centre.
    Step 1: Alternate segment theorem.
    ∠PTM=∠PQT=34∘and∠STN=∠SRT=56∘\angle PTM = \angle PQT = 34^{\circ} \quad \text{and} \quad \angle STN = \angle SRT = 56^{\circ}
    (Reason: ∠PTM\angle PTM is the tangent-chord angle for chord TPTP, so it equals ∠PQT\angle PQT in the alternate segment; similarly ∠STN=∠SRT\angle STN = \angle SRT.)
    Step 2: Angles on a straight line.
    ∠PTS=180∘−34∘−56∘=90∘\angle PTS = 180^{\circ} - 34^{\circ} - 56^{\circ} = 90^{\circ}
    (Reason: MTNMTN is a straight line, so ∠PTM+∠PTS+∠STN=180∘\angle PTM + \angle PTS + \angle STN = 180^{\circ}.)
    Step 3: Angle in a semicircle (converse).
    ∠PTS=90∘⇒PS is a diameter\angle PTS = 90^{\circ} \Rightarrow PS \text{ is a diameter}
    (Reason: The angle in a semicircle is 90∘90^{\circ}; conversely, if an inscribed angle is 90∘90^{\circ} then the chord opposite it is a diameter, which passes through the centre.)
    Therefore PSPS is a diameter, so it passes through the centre of the circle.
    Verification
    Check 1: Angles on the straight line at TT: 34∘+90∘+56∘=180∘34^{\circ} + 90^{\circ} + 56^{\circ} = 180^{\circ}.
    Check 2: The two given angles satisfy 34∘+56∘=90∘34^{\circ} + 56^{\circ} = 90^{\circ}, which is why ∠PTS=180∘−90∘=90∘\angle PTS = 180^{\circ} - 90^{\circ} = 90^{\circ}.
    Check 3: ∠PTS=90∘\angle PTS = 90^{\circ} is the angle in a semicircle, confirming PSPS is a diameter.
    Mark Scheme Breakdown
    StepMarkDescriptionGot it?
    ∠PTM=34∘ and ∠STN=56∘\angle PTM = 34^{\circ} \text{ and } \angle STN = 56^{\circ}M1Method - alternate segment theorem✓
    ∠PTS=90∘\angle PTS = 90^{\circ}M1Method - angles on a straight line✓
    PSPS is a diameter through the centreA1Accuracy - angle in a semicircle, so PS is a diameter✓

    Full marks: 3/3

    Question 6, Grade 8, Non-calculator

    Circle centre O with tangent MN at S; radius OR perpendicular to chord SQ; angle QSN = 58 degreesSQRPOMN58°Not drawn accurately

    PP, QQ, RR and SS lie on a circle, centre OO. MNMN is the tangent at SS. ∠QSN=58∘\angle QSN = 58^{\circ}.

    (a) Work out the size of ∠SOQ\angle SOQ. [2 marks]

    (b) Explain why ∠ROQ\angle ROQ is half the size of ∠SOQ\angle SOQ. [2 marks]

    (a)∠SOQ=\angle SOQ =∘^{\circ}
    [Total 4 marks]
    GRADE8
    Show solution & mark schemeHide solution & mark scheme

    Question 6 - Exam Solution

    Understanding the Question
    Given
    P,Q,R,SP, Q, R, S lie on a circle, centre OO; MNMN is the tangent at SS.
    ∠QSN=58∘\angle QSN = 58^{\circ}.
    Find
    (a) The size of ∠SOQ\angle SOQ.
    (b) Why ∠ROQ\angle ROQ is half the size of ∠SOQ\angle SOQ.
    Plan the Solution
    • Part (a): alternate segment theorem, then the centre-to-circumference link.
    • Part (b): the radius perpendicular to a chord bisects both the chord and the angle at the centre.
    Part (a): Work out ∠SOQ\angle SOQ  [2 marks]
    Rule - Alternate segment, then double at the centre: transfer the tangent-chord angle to the circumference, then use the centre-to-circumference link.
    Step 1: Alternate segment theorem.
    ∠QPS=∠QSN=58∘\angle QPS = \angle QSN = 58^{\circ}
    (Reason: ∠QSN\angle QSN is the tangent-chord angle for chord SQSQ, equal to ∠QPS\angle QPS in the alternate segment.)
    Step 2: Angle at the centre is twice the angle at the circumference.
    ∠SOQ=2×58∘=116∘\angle SOQ = 2 \times 58^{\circ} = 116^{\circ}
    (Reason: ∠SOQ\angle SOQ is at the centre and ∠QPS\angle QPS is at the circumference, both standing on arc SQSQ.)
    ∠SOQ=116∘\angle SOQ = 116^{\circ}
    Mark scheme
    StepMarkDescriptionGot it?
    ∠QPS=58∘\angle QPS = 58^{\circ}M1Method - alternate segment theorem✓
    ∠SOQ=116∘\angle SOQ = 116^{\circ}A1Accuracy - angle at centre is twice the circumference✓

    Full marks: 2/2

    Part (b): Explain why ∠ROQ=12∠SOQ\angle ROQ = \tfrac{1}{2}\angle SOQ  [2 marks]
    Rule - Isosceles symmetry: the radius that is perpendicular to a chord bisects both the chord and the angle at the centre.
    Step 1: Isosceles triangle and perpendicular bisector.
    OQ=OSOQ = OS (radii), so triangle OSQOSQ is isosceles. OROR meets chord SQSQ at right angles, so it bisects SQSQ.
    (Reason: All radii are equal, and the perpendicular from the centre of a circle to a chord bisects the chord.)
    Step 2: Bisecting the base bisects the angle at O.
    Bisecting the base of the isosceles triangle also bisects the apex angle, so ∠ROQ=12×∠SOQ\angle ROQ = \tfrac{1}{2} \times \angle SOQ.
    (Reason: In an isosceles triangle, the line from the apex perpendicular to the base bisects the apex angle.)
    Therefore ∠ROQ=12×∠SOQ\angle ROQ = \tfrac{1}{2} \times \angle SOQ.
    Mark scheme
    StepMarkDescriptionGot it?
    Triangle OSQOSQ isosceles and OROR bisects SQSQM1Reason - perpendicular from the centre bisects a chord✓
    ∠ROQ=12∠SOQ\angle ROQ = \tfrac{1}{2}\angle SOQA1Reason - the bisector halves the apex angle✓

    Full marks: 2/2

    Verification
    Check 1 (a): ∠SOQ=2×∠QPS=2×58∘=116∘\angle SOQ = 2 \times \angle QPS = 2 \times 58^{\circ} = 116^{\circ}
    Check 2 (b): OROR splits triangle OSQOSQ symmetrically, so ∠ROQ=∠SOR=12×116∘=58∘\angle ROQ = \angle SOR = \tfrac{1}{2} \times 116^{\circ} = 58^{\circ}, which is half of ∠SOQ\angle SOQ.

    Question 7, Grade 9, Non-calculator

    Circle with points P, Q, R, S and interior point Y (near the centre) joined to S and R; angle PQR = 118, angle YSP = 29, angle YRS = 31 degreesPQRSY118°29°31°Not drawn accurately

    PP, QQ, RR and SS lie on a circle. YY is a point inside the circle, joined to SS and RR. ∠PQR=118∘\angle PQR = 118^{\circ}, ∠YSP=29∘\angle YSP = 29^{\circ} and ∠YRS=31∘\angle YRS = 31^{\circ}. Show that YY is not the centre of the circle.

    [Total 3 marks]
    GRADE9Problem solving
    Show solution & mark schemeHide solution & mark scheme

    Question 7 - Exam Solution

    Understanding the Question
    Given
    P,Q,R,SP, Q, R, S lie on a circle. YY is inside, joined to SS and RR.
    ∠PQR=118∘\angle PQR = 118^{\circ}, ∠YSP=29∘\angle YSP = 29^{\circ}, ∠YRS=31∘\angle YRS = 31^{\circ}.
    Show
    That YY is not the centre of the circle.
    Plan the Solution
    • Cyclic quadrilateral PQRSPQRS: find ∠PSR\angle PSR.
    • Subtract to find ∠RSY\angle RSY.
    • If YY were the centre, YSYS and YRYR would be radii, giving equal base angles. Check whether ∠RSY=∠SRY\angle RSY = \angle SRY.
    Worked Solution [3 marks]
    Rule - Assume and contradict: suppose YY is the centre; two radii would force equal base angles, so if the angles differ, YY cannot be the centre.
    Step 1: Opposite angles of a cyclic quadrilateral.
    ∠PSR=180∘−118∘=62∘\angle PSR = 180^{\circ} - 118^{\circ} = 62^{\circ}
    (Reason: PQRSPQRS is a cyclic quadrilateral, so the opposite angles ∠PQR\angle PQR and ∠PSR\angle PSR add up to 180∘180^{\circ}.)
    Step 2: Subtract to find angle RSY.
    ∠RSY=∠PSR−∠PSY=62∘−29∘=33∘\angle RSY = \angle PSR - \angle PSY = 62^{\circ} - 29^{\circ} = 33^{\circ}
    (Reason: ∠PSY\angle PSY is part of ∠PSR\angle PSR, so subtracting it leaves ∠RSY\angle RSY.)
    Step 3: Test the "centre" assumption.
    If YY were the centre, YSYS and YRYR would be radii, so triangle RSYRSY would be isosceles with ∠RSY=∠SRY\angle RSY = \angle SRY. But ∠RSY=33∘\angle RSY = 33^{\circ} and ∠SRY=31∘\angle SRY = 31^{\circ}.
    (Reason: A triangle made from two radii is isosceles, so its base angles would be equal - but here they are not.)
    Since ∠RSY=33∘≠31∘=∠SRY\angle RSY = 33^{\circ} \ne 31^{\circ} = \angle SRY, YY is not the centre of the circle.
    Verification
    Check 1: Cyclic quadrilateral: ∠PQR+∠PSR=118∘+62∘=180∘\angle PQR + \angle PSR = 118^{\circ} + 62^{\circ} = 180^{\circ}.
    Check 2: The parts add up: ∠PSY+∠RSY=29∘+33∘=62∘=∠PSR\angle PSY + \angle RSY = 29^{\circ} + 33^{\circ} = 62^{\circ} = \angle PSR.
    Check 3: Two radii would force ∠RSY=∠SRY\angle RSY = \angle SRY, but 33∘≠31∘33^{\circ} \ne 31^{\circ}, confirming YY is not the centre.
    Mark Scheme Breakdown
    StepMarkDescriptionGot it?
    ∠PSR=62∘\angle PSR = 62^{\circ}M1Reason - opposite angles in a cyclic quadrilateral✓
    ∠RSY=33∘\angle RSY = 33^{\circ}M1Method - angle subtraction✓
    33∘≠31∘33^{\circ} \ne 31^{\circ}, so YY is not the centreA1Reason - two radii would give equal base angles✓

    Full marks: 3/3

    Question 8, Grade 8, Non-calculator

    Equilateral triangle PQR in a circle with PQ = PR and tangent ST at R parallel to PQ; angle SRP = xPQRSTxNot drawn accurately

    PP, QQ and RR lie on a circle with PQ=PRPQ = PR. STST is the tangent at RR, and STST is parallel to PQPQ. ∠SRP=x\angle SRP = x. Prove that triangle PQRPQR is an equilateral triangle.

    [Total 4 marks]
    GRADE8Problem solving
    Show solution & mark schemeHide solution & mark scheme

    Question 8 - Exam Solution

    Understanding the Question
    Given
    Triangle PQRPQR with P,Q,RP, Q, R on a circle; PQ=PRPQ = PR.
    STST is a tangent at RR, parallel to PQPQ; ∠SRP=x\angle SRP = x.
    Prove
    That triangle PQRPQR is an equilateral triangle.
    Plan the Solution
    • Alternate segment theorem: ∠PQR=x\angle PQR = x.
    • Isosceles triangle (PQ=PRPQ = PR): ∠PRQ=x\angle PRQ = x.
    • Alternate angles (STST parallel to PQPQ): ∠QPR=x\angle QPR = x.
    • All three angles equal, so the triangle is equilateral.
    Worked Solution [4 marks]
    Rule - Show all three angles are equal: a triangle is equilateral when its three angles are equal; find each angle in turn and show they are all xx.
    Step 1: Alternate segment theorem.
    ∠PQR=x\angle PQR = x
    (Reason: ∠SRP\angle SRP is the angle between the tangent STST and the chord RPRP, so by the alternate segment theorem it equals ∠PQR\angle PQR in the alternate segment.)
    Step 2: Isosceles triangle.
    ∠PRQ=x\angle PRQ = x
    (Reason: PQ=PRPQ = PR, so triangle PQRPQR is isosceles; its base angles are equal, so ∠PRQ=∠PQR=x\angle PRQ = \angle PQR = x.)
    Step 3: Alternate angles.
    ∠QPR=x\angle QPR = x
    (Reason: STST is parallel to PQPQ, so ∠SRP\angle SRP and ∠QPR\angle QPR are alternate angles and are equal.)
    Step 4: All three angles are equal.
    ∠PQR=∠PRQ=∠QPR=x\angle PQR = \angle PRQ = \angle QPR = x
    (Reason: A triangle with three equal angles is equilateral, so each angle is 60∘60^{\circ}.)
    All three angles equal xx, so triangle PQRPQR is equilateral (each angle is 60∘60^{\circ}).
    Verification
    Check 1: The three equal angles sum to 180∘180^{\circ}: x+x+x=3x=180∘x + x + x = 3x = 180^{\circ}, so x=60∘x = 60^{\circ}.
    Check 2: Equal angles give equal sides, so PQ=QR=RPPQ = QR = RP, which is exactly what equilateral means.
    Check 3: With x=60∘x = 60^{\circ}, the tangent-chord angle is 60∘60^{\circ}, consistent with an equilateral triangle inscribed in the circle.
    Mark Scheme Breakdown
    StepMarkDescriptionGot it?
    ∠PQR=x\angle PQR = xM1Reason - alternate segment theorem✓
    ∠PRQ=x\angle PRQ = xM1Reason - isosceles triangle (PQ = PR)✓
    ∠QPR=x\angle QPR = xM1Reason - alternate angles (ST parallel to PQ)✓
    All three angles equal, so equilateralA1Reason - three equal angles mean equilateral✓

    Full marks: 4/4

    Question 9, Grade 8, Non-calculator

    Circle centre O with points P, Q, R, S and radii; angle POS = 118, angle PQO = 56, angle ORQ = 42 degreesPQRSO118°56°42°Not drawn accurately

    The diagram shows a circle, centre OO. PP, QQ, RR and SS lie on the circle. ∠POS=118∘\angle POS = 118^{\circ}, ∠PQO=56∘\angle PQO = 56^{\circ} and ∠ORQ=42∘\angle ORQ = 42^{\circ}. Find the size of ∠RSO\angle RSO.

    ∠RSO=\angle RSO =∘^{\circ}
    [Total 4 marks]
    GRADE8
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    Question 9 - Exam Solution

    Understanding the Question
    Given
    Circle, centre OO; P,Q,R,SP, Q, R, S on the circle.
    ∠POS=118∘\angle POS = 118^{\circ}, ∠PQO=56∘\angle PQO = 56^{\circ}, ∠ORQ=42∘\angle ORQ = 42^{\circ}.
    Find
    The size of ∠RSO\angle RSO.
    Plan the Solution
    • Triangle POSPOS is isosceles (two radii): find ∠OSP\angle OSP.
    • Triangle QORQOR is isosceles (two radii): find ∠OQR\angle OQR.
    • Combine at QQ to get ∠PQR\angle PQR, then use the cyclic quadrilateral for ∠RSP\angle RSP.
    • Subtract to find ∠RSO\angle RSO.
    Worked Solution [4 marks]
    Rule - Radii make isosceles triangles: every pair of radii forms an isosceles triangle with equal base angles; use these, then the cyclic-quadrilateral rule, and subtract.
    Step 1: Isosceles triangle POS.
    ∠OSP=12(180∘−118∘)=12×62∘=31∘\angle OSP = \tfrac{1}{2}(180^{\circ} - 118^{\circ}) = \tfrac{1}{2} \times 62^{\circ} = 31^{\circ}
    (Reason: OP=OSOP = OS are radii, so triangle POSPOS is isosceles; the base angles are equal, each being half of what is left after the apex angle.)
    Step 2: Isosceles triangle QOR.
    ∠OQR=∠ORQ=42∘\angle OQR = \angle ORQ = 42^{\circ}
    (Reason: OQ=OROQ = OR are radii, so triangle QORQOR is isosceles; its base angles are equal.)
    Step 3: Cyclic quadrilateral PQRS.
    ∠PQR=∠PQO+∠OQR=56∘+42∘=98∘\angle PQR = \angle PQO + \angle OQR = 56^{\circ} + 42^{\circ} = 98^{\circ}
    ∠RSP=180∘−98∘=82∘\angle RSP = 180^{\circ} - 98^{\circ} = 82^{\circ}
    (Reason: Adding the two parts at QQ gives ∠PQR\angle PQR; then PQRSPQRS is a cyclic quadrilateral, so the opposite angles ∠PQR\angle PQR and ∠RSP\angle RSP add up to 180∘180^{\circ}.)
    Step 4: Subtract to find angle RSO.
    ∠RSO=∠RSP−∠OSP=82∘−31∘=51∘\angle RSO = \angle RSP - \angle OSP = 82^{\circ} - 31^{\circ} = 51^{\circ}
    (Reason: ∠OSP\angle OSP is part of ∠RSP\angle RSP, so subtracting it leaves ∠RSO\angle RSO.)
    ∠RSO=51∘\angle RSO = 51^{\circ}
    Verification
    Check 1: Triangle POSPOS: 118∘+31∘+31∘=180∘118^{\circ} + 31^{\circ} + 31^{\circ} = 180^{\circ}.
    Check 2: Cyclic quadrilateral: ∠PQR+∠RSP=98∘+82∘=180∘\angle PQR + \angle RSP = 98^{\circ} + 82^{\circ} = 180^{\circ}.
    Check 3: The parts add up: ∠RSO+∠OSP=51∘+31∘=82∘=∠RSP\angle RSO + \angle OSP = 51^{\circ} + 31^{\circ} = 82^{\circ} = \angle RSP.
    Mark Scheme Breakdown
    StepMarkDescriptionGot it?
    ∠OSP=31∘\angle OSP = 31^{\circ}M1Method - isosceles triangle (two radii)✓
    ∠OQR=42∘\angle OQR = 42^{\circ}M1Method - isosceles triangle (two radii)✓
    ∠RSP=82∘\angle RSP = 82^{\circ}M1Method - cyclic quadrilateral opposite angles✓
    ∠RSO=51∘\angle RSO = 51^{\circ}A1Accuracy - correct final answer✓

    Full marks: 4/4

    Question 10, Grade 8, Non-calculator

    Circle centre O with diameter PQ and point R on the circumference; triangle PRQ drawnPQRONot drawn accurately

    The diagram shows a circle with centre OO. PQPQ is a diameter and RR is a point on the circumference. Prove that ∠PRQ=90∘\angle PRQ = 90^{\circ}.

    [Total 3 marks]
    GRADE8Problem solving
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    Question 10 - Exam Solution

    Understanding the Question
    Given
    A circle with centre OO. PQPQ is a diameter and RR is a point on the circumference.
    Prove
    That ∠PRQ=90∘\angle PRQ = 90^{\circ}.
    Plan the Solution
    • Draw the radius OROR, splitting triangle PQRPQR into two isosceles triangles.
    • Label the base angles xx and yy.
    • Use the angle sum of triangle PQRPQR to show 2x+2y=180∘2x + 2y = 180^{\circ}, hence x+y=90∘x + y = 90^{\circ}.
    Worked Solution [3 marks]
    Rule - Two radii, two isosceles triangles: the radius to RR creates two isosceles triangles; adding their base angles and using the triangle angle sum forces the angle at RR to be 90∘90^{\circ}.
    Step 1: Split into two isosceles triangles.
    Draw OROR. Since OP=OROP = OR and OQ=OROQ = OR (radii):
    ∠OPR=∠ORP=xand∠OQR=∠ORQ=y\angle OPR = \angle ORP = x \quad \text{and} \quad \angle OQR = \angle ORQ = y
    Answer diagram: radius OR splits triangle PQR into two isosceles triangles with base angles x, x and y, yPQROxyxyNot drawn accurately
    (Reason: OPOP, OQOQ and OROR are all radii, so triangles OPROPR and OQROQR are isosceles, with equal base angles.)
    Step 2: Angle sum of triangle PQR.
    The angle at RR is ∠PRQ=x+y\angle PRQ = x + y, so:
    x+x+y+y=180∘x + x + y + y = 180^{\circ}
    (Reason: The angles in triangle PQRPQR - namely xx, yy and x+yx + y - add up to 180∘180^{\circ}.)
    Step 3: Solve for the angle at R.
    2x+2y=180∘, so x+y=90∘2x + 2y = 180^{\circ}, \text{ so } x + y = 90^{\circ}
    ∠PRQ=x+y=90∘\angle PRQ = x + y = 90^{\circ}
    (Reason: Dividing by 2 gives x+y=90∘x + y = 90^{\circ}, and ∠PRQ\angle PRQ equals x+yx + y.)
    Therefore ∠PRQ=90∘\angle PRQ = 90^{\circ} (the angle in a semicircle is a right angle).
    Verification
    Check 1: x+x+y+y=2(x+y)=2×90∘=180∘x + x + y + y = 2(x + y) = 2 \times 90^{\circ} = 180^{\circ}, matching the triangle angle sum.
    Check 2: The proof uses no fixed value for xx or yy, so ∠PRQ=90∘\angle PRQ = 90^{\circ} for any point RR on the semicircle.
    Check 3: A right angle at RR means the diameter PQPQ subtends 90∘90^{\circ}, which is exactly the angle in a semicircle.
    Mark Scheme Breakdown
    StepMarkDescriptionGot it?
    Two isosceles triangles, base angles xx and yyM1Method - split using the radius OR✓
    x+x+y+y=180∘x + x + y + y = 180^{\circ}M1Method - angles in a triangle✓
    x+y=90∘x + y = 90^{\circ}, so ∠PRQ=90∘\angle PRQ = 90^{\circ}A1Reason - the angle in a semicircle is 90 degrees✓

    Full marks: 3/3

    Question 11, Grade 9, Non-calculator

    Two intersecting circles centres O and P meeting at G and H, with points W, X, Y, Z; angle GWY = 74, angle GPH = 84WXYGHZOP74°84°Not drawn accurately

    The diagram shows two intersecting circles with centres OO and PP. The circles intersect at GG and HH. WW, XX and YY are points on the circle with centre OO, and ZZ is a point on the circle with centre PP. XZXZ and ZYZY are straight lines. ∠GWY=74∘\angle GWY = 74^{\circ} and ∠GPH=84∘\angle GPH = 84^{\circ}. Find the size of ∠XYH\angle XYH, giving reasons.

    ∠XYH=\angle XYH =∘^{\circ}
    [Total 3 marks]
    GRADE9
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    Question 11 - Exam Solution

    Understanding the Question
    Given
    Two intersecting circles, centres OO and PP, meeting at GG and HH. W,X,YW, X, Y on circle OO; ZZ on circle PP. XZXZ and ZYZY are straight lines.
    ∠GWY=74∘\angle GWY = 74^{\circ} and ∠GPH=84∘\angle GPH = 84^{\circ}.
    Find
    The size of ∠XYH\angle XYH, giving reasons.
    Plan the Solution
    • Circle PP: ∠GPH\angle GPH is at the centre, so the inscribed angle ∠GZH\angle GZH is half of it.
    • Circle OO: ∠YXG\angle YXG equals ∠GWY\angle GWY (same segment).
    • These are two angles of triangle XZYXZY (as GG lies on XZXZ and HH lies on ZYZY). Use the triangle angle sum for ∠XYH\angle XYH.
    Worked Solution [3 marks]
    Rule - Work through each circle: use the centre-to-circumference rule in circle PP and the same-segment rule in circle OO to fill in two angles of triangle XZYXZY, then use the angle sum.
    Step 1: Circle P - angle at the centre.
    ∠GZH=12×∠GPH=12×84∘=42∘\angle GZH = \tfrac{1}{2} \times \angle GPH = \tfrac{1}{2} \times 84^{\circ} = 42^{\circ}
    (Reason: In circle PP, ∠GPH\angle GPH is at the centre and ∠GZH\angle GZH is at the circumference, both standing on arc GHGH.)
    Step 2: Circle O - angles in the same segment.
    ∠YXG=∠GWY=74∘\angle YXG = \angle GWY = 74^{\circ}
    (Reason: In circle OO, ∠YXG\angle YXG and ∠GWY\angle GWY both stand on the chord GYGY, so angles in the same segment are equal.)
    Step 3: Angles in triangle XZY.
    GG lies on XZXZ and HH lies on ZYZY, so ∠XZY=∠GZH=42∘\angle XZY = \angle GZH = 42^{\circ} and ∠XYH=∠XYZ\angle XYH = \angle XYZ.
    ∠XYH=180∘−74∘−42∘=64∘\angle XYH = 180^{\circ} - 74^{\circ} - 42^{\circ} = 64^{\circ}
    (Reason: The angles in triangle XZYXZY add up to 180∘180^{\circ}; ∠XYH\angle XYH is the angle at YY because HH lies on ZYZY.)
    ∠XYH=64∘\angle XYH = 64^{\circ}
    Verification
    Check 1: Triangle XZYXZY: 74∘+42∘+64∘=180∘74^{\circ} + 42^{\circ} + 64^{\circ} = 180^{\circ}.
    Check 2: ∠GZH=42∘\angle GZH = 42^{\circ} is exactly half of ∠GPH=84∘\angle GPH = 84^{\circ} (centre is twice the circumference).
    Check 3: ∠YXG=∠GWY=74∘\angle YXG = \angle GWY = 74^{\circ}, confirming the same-segment step.
    Mark Scheme Breakdown
    StepMarkDescriptionGot it?
    ∠GZH=42∘\angle GZH = 42^{\circ}M1Method - angle at the centre is twice the circumference (circle P)✓
    ∠YXG=74∘\angle YXG = 74^{\circ}M1Method - angles in the same segment (circle O)✓
    ∠XYH=64∘\angle XYH = 64^{\circ}A1Accuracy - angles in a triangle✓

    Full marks: 3/3

    Question 12, Grade 9, Non-calculator

    Circle with chords PR and QS meeting at T near the centre; MN tangent at S; angle QTR = 70 and angle NSR = 58 degreesPQRSTMN70°58°Not drawn accurately

    PP, QQ, RR and SS lie on a circle. MNMN is a tangent at SS, and PRPR and QSQS are straight lines meeting at TT. ∠QTR=70∘\angle QTR = 70^{\circ} and ∠NSR=58∘\angle NSR = 58^{\circ}. Show that TT is not the centre of the circle.

    [Total 4 marks]
    GRADE9Problem solving
    Show solution & mark schemeHide solution & mark scheme

    Question 12 - Exam Solution

    Understanding the Question
    Given
    P,Q,R,SP, Q, R, S lie on a circle. MNMN is a tangent at SS; PRPR and QSQS are straight lines meeting at TT.
    ∠QTR=70∘\angle QTR = 70^{\circ} and ∠NSR=58∘\angle NSR = 58^{\circ}.
    Show
    That TT is not the centre of the circle.
    Plan the Solution
    • Alternate segment theorem: find ∠SQR\angle SQR.
    • Angles on a straight line at TT: find ∠STR\angle STR.
    • If TT were the centre, ∠STR\angle STR would equal 2×∠SQR2 \times \angle SQR. Check whether this holds.
    Worked Solution [4 marks]
    Rule - Assume and contradict: suppose TT is the centre; the angle at the centre would be twice the inscribed angle, so if that fails, TT cannot be the centre.
    Step 1: Alternate segment theorem.
    ∠SQR=∠NSR=58∘\angle SQR = \angle NSR = 58^{\circ}
    (Reason: ∠NSR\angle NSR is the tangent-chord angle for chord SRSR, equal to ∠SQR\angle SQR in the alternate segment.)
    Step 2: Angles on a straight line at T.
    ∠STR=180∘−70∘=110∘\angle STR = 180^{\circ} - 70^{\circ} = 110^{\circ}
    (Reason: QSQS is a straight line, so ∠QTR\angle QTR and ∠STR\angle STR add up to 180∘180^{\circ}.)
    Step 3: Apply the "centre" assumption.
    If TT were the centre, TSTS and TRTR would be radii, and:
    ∠STR=2×∠SQR=2×58∘=116∘\angle STR = 2 \times \angle SQR = 2 \times 58^{\circ} = 116^{\circ}
    (Reason: The angle at the centre is twice the angle at the circumference standing on the same arc SRSR.)
    Step 4: Compare.
    But ∠STR=110∘≠116∘.\text{But } \angle STR = 110^{\circ} \ne 116^{\circ}.
    (Reason: The actual angle at TT does not match the value it would have if TT were the centre.)
    Since ∠STR=110∘≠116∘=2×∠SQR\angle STR = 110^{\circ} \ne 116^{\circ} = 2 \times \angle SQR, TT is not the centre of the circle.
    Verification
    Check 1: Straight line at TT: ∠QTR+∠STR=70∘+110∘=180∘\angle QTR + \angle STR = 70^{\circ} + 110^{\circ} = 180^{\circ}.
    Check 2: If TT were the centre, ∠STR=2×58∘=116∘\angle STR = 2 \times 58^{\circ} = 116^{\circ}, but the straight line gives 110∘110^{\circ} - the two disagree.
    Check 3: 110∘≠116∘110^{\circ} \ne 116^{\circ} confirms TT cannot be the centre.
    Mark Scheme Breakdown
    StepMarkDescriptionGot it?
    ∠SQR=58∘\angle SQR = 58^{\circ}M1Reason - alternate segment theorem✓
    ∠STR=110∘\angle STR = 110^{\circ}M1Method - angles on a straight line✓
    If centre, ∠STR=116∘\angle STR = 116^{\circ}M1Reason - angle at centre is twice the circumference✓
    116∘≠110∘116^{\circ} \ne 110^{\circ}, so TT is not the centreA1Reason - the values disagree✓

    Full marks: 4/4

    Question 13, Grade 9, Non-calculator

    Circle centre O with tangents JP and JS from J and straight lines PK and SK meeting at K; angle PJS = 40, angle OPQ = 52, angle PQR = 116JPQRSKO40°52°116°Not drawn accurately

    The diagram shows a circle, centre OO. PP, QQ, RR and SS lie on the circle. JPJP and JSJS are tangents from JJ. PKPK and SKSK are straight lines. ∠PJS=40∘\angle PJS = 40^{\circ}, ∠OPQ=52∘\angle OPQ = 52^{\circ} and ∠PQR=116∘\angle PQR = 116^{\circ}. Find the size of ∠QKR\angle QKR.

    ∠QKR=\angle QKR =∘^{\circ}
    [Total 5 marks]
    GRADE9
    Show solution & mark schemeHide solution & mark scheme

    Question 13 - Exam Solution

    Understanding the Question
    Given
    Circle, centre OO; P,Q,R,SP, Q, R, S on the circle. JPJP and JSJS are tangents from JJ; PKPK and SKSK are straight lines.
    ∠PJS=40∘\angle PJS = 40^{\circ}, ∠OPQ=52∘\angle OPQ = 52^{\circ}, ∠PQR=116∘\angle PQR = 116^{\circ}.
    Find
    The size of ∠QKR\angle QKR.
    Plan the Solution
    • Tangents from JJ are equal, so triangle JPSJPS is isosceles: find the base angles.
    • Tangent meets radius, then the isosceles triangle OPSOPS: find ∠SPO\angle SPO.
    • Add at PP to get ∠SPQ\angle SPQ, then use the cyclic quadrilateral for ∠SRQ\angle SRQ.
    • Angles on the straight lines PKPK and SKSK: find ∠KQR\angle KQR and ∠KRQ\angle KRQ.
    • Angles in triangle QKRQKR: find ∠QKR\angle QKR.
    Worked Solution [5 marks]
    Rule - Work from the tangents to the triangle: build up the angles from the two tangents, use the centre and the cyclic quadrilateral, then finish with the triangle at KK.
    Step 1: Tangents from J are equal (isosceles triangle).
    ∠JPS=∠JSP=12(180∘−40∘)=12×140∘=70∘\angle JPS = \angle JSP = \tfrac{1}{2}(180^{\circ} - 40^{\circ}) = \tfrac{1}{2} \times 140^{\circ} = 70^{\circ}
    (Reason: JPJP and JSJS are equal tangents from JJ, so triangle JPSJPS is isosceles with equal base angles.)
    Step 2: Tangent-radius, then isosceles triangle OPS.
    ∠JSO=90∘, so ∠PSO=90∘−70∘=20∘, and ∠SPO=∠PSO=20∘\angle JSO = 90^{\circ}, \text{ so } \angle PSO = 90^{\circ} - 70^{\circ} = 20^{\circ}, \text{ and } \angle SPO = \angle PSO = 20^{\circ}
    (Reason: OSOS is a radius and JSJS a tangent, so ∠JSO=90∘\angle JSO = 90^{\circ}; and OP=OSOP = OS are radii, so triangle OPSOPS is isosceles with equal base angles.)
    Step 3: Cyclic quadrilateral PQRS.
    ∠SPQ=∠SPO+∠OPQ=20∘+52∘=72∘\angle SPQ = \angle SPO + \angle OPQ = 20^{\circ} + 52^{\circ} = 72^{\circ}
    ∠SRQ=180∘−72∘=108∘\angle SRQ = 180^{\circ} - 72^{\circ} = 108^{\circ}
    (Reason: Adding the parts at PP gives ∠SPQ\angle SPQ; then PQRSPQRS is a cyclic quadrilateral, so opposite angles ∠SPQ\angle SPQ and ∠SRQ\angle SRQ add up to 180∘180^{\circ}.)
    Step 4: Angles on the straight lines.
    ∠KQR=180∘−116∘=64∘and∠KRQ=180∘−108∘=72∘\angle KQR = 180^{\circ} - 116^{\circ} = 64^{\circ} \quad \text{and} \quad \angle KRQ = 180^{\circ} - 108^{\circ} = 72^{\circ}
    (Reason: PQKPQK and SRKSRK are straight lines, so the angles at QQ and at RR each add up to 180∘180^{\circ}.)
    Step 5: Angles in triangle QKR.
    ∠QKR=180∘−64∘−72∘=44∘\angle QKR = 180^{\circ} - 64^{\circ} - 72^{\circ} = 44^{\circ}
    (Reason: The angles in triangle QKRQKR add up to 180∘180^{\circ}.)
    ∠QKR=44∘\angle QKR = 44^{\circ}
    Verification
    Check 1: Triangle JPSJPS: 40∘+70∘+70∘=180∘40^{\circ} + 70^{\circ} + 70^{\circ} = 180^{\circ}.
    Check 2: Cyclic quadrilateral: ∠SPQ+∠SRQ=72∘+108∘=180∘\angle SPQ + \angle SRQ = 72^{\circ} + 108^{\circ} = 180^{\circ}.
    Check 3: Triangle QKRQKR: 64∘+72∘+44∘=180∘64^{\circ} + 72^{\circ} + 44^{\circ} = 180^{\circ}.
    Mark Scheme Breakdown
    StepMarkDescriptionGot it?
    ∠JPS=∠JSP=70∘\angle JPS = \angle JSP = 70^{\circ}M1Method - tangents from a point are equal (isosceles)✓
    ∠SPO=20∘\angle SPO = 20^{\circ}M1Method - tangent-radius and isosceles triangle✓
    ∠SRQ=108∘\angle SRQ = 108^{\circ}M1Method - cyclic quadrilateral opposite angles✓
    ∠KQR=64∘,∠KRQ=72∘\angle KQR = 64^{\circ}, \angle KRQ = 72^{\circ}M1Method - angles on straight lines✓
    ∠QKR=44∘\angle QKR = 44^{\circ}A1Accuracy - angles in a triangle✓

    Full marks: 5/5

    Question 14, Grade 9, Non-calculator

    Circle centre O with tangent JK at P and tangent ST at R and triangle PQR; angles x, y and b markedPQRSTJKOxybNot drawn accurately

    The diagram shows a circle, centre OO. JKJK is a tangent at PP and STST is a tangent at RR. Here xx is the tangent-chord angle at PP, yy is the tangent-chord angle at RR, and b=∠ORQb = \angle ORQ. Prove that b=x+y−90∘b = x + y - 90^{\circ}. State any circle theorems that you use.

    [Total 4 marks]
    GRADE9Problem solving
    Show solution & mark schemeHide solution & mark scheme

    Question 14 - Exam Solution

    Understanding the Question
    Given
    Circle, centre OO. JKJK is a tangent at PP and STST is a tangent at RR.
    xx is the tangent-chord angle at PP, yy the tangent-chord angle at RR, and b=∠ORQb = \angle ORQ.
    Prove
    That b=x+y−90∘b = x + y - 90^{\circ}.
    Plan the Solution
    • Alternate segment theorem: ∠PRQ=x\angle PRQ = x.
    • Tangent meets radius at RR: ∠ORS=90∘\angle ORS = 90^{\circ}.
    • Subtract to find ∠ORP=90∘−y\angle ORP = 90^{\circ} - y.
    • bb is ∠PRQ\angle PRQ minus ∠ORP\angle ORP; simplify to get the result.
    Worked Solution [4 marks]
    Rule - Build the target from known angles: write bb as a difference of angles found with the alternate segment theorem and the tangent-radius right angle, then simplify.
    Step 1: Alternate segment theorem.
    ∠PRQ=x\angle PRQ = x
    (Reason: xx is the angle between the tangent at PP and the chord PQPQ, so by the alternate segment theorem it equals ∠PRQ\angle PRQ in the alternate segment.)
    Step 2: Tangent meets radius.
    ∠ORS=90∘\angle ORS = 90^{\circ}
    (Reason: OROR is a radius and RSRS is a tangent; a tangent meets a radius at 90∘90^{\circ}.)
    Step 3: Subtract to find angle ORP.
    ∠ORP=∠ORS−∠SRP=90∘−y\angle ORP = \angle ORS - \angle SRP = 90^{\circ} - y
    (Reason: y=∠SRPy = \angle SRP is part of ∠ORS\angle ORS, so subtracting it gives ∠ORP\angle ORP.)
    Step 4: Form b and simplify.
    b=∠PRQ−∠ORP=x−(90∘−y)=x+y−90∘b = \angle PRQ - \angle ORP = x - (90^{\circ} - y) = x + y - 90^{\circ}
    (Reason: bb is the part of ∠PRQ\angle PRQ that remains after ∠ORP\angle ORP is removed; expanding the bracket gives the result.)
    Therefore b=x+y−90∘b = x + y - 90^{\circ}, as required.
    Verification
    Check 1: For example, if x=55∘x = 55^{\circ} and y=60∘y = 60^{\circ}, the formula gives b=55∘+60∘−90∘=25∘b = 55^{\circ} + 60^{\circ} - 90^{\circ} = 25^{\circ}, which agrees with the actual ∠ORQ\angle ORQ.
    Check 2: Rearranging gives x+y=b+90∘x + y = b + 90^{\circ}, so the two tangent-chord angles together are always 90∘90^{\circ} more than bb.
    Mark Scheme Breakdown
    StepMarkDescriptionGot it?
    ∠PRQ=x\angle PRQ = xM1Reason - alternate segment theorem✓
    ∠ORS=90∘\angle ORS = 90^{\circ}M1Reason - tangent meets a radius at 90 degrees✓
    ∠ORP=90∘−y\angle ORP = 90^{\circ} - yM1Method - angle subtraction✓
    b=x+y−90∘b = x + y - 90^{\circ}A1Accuracy - correct algebraic result✓

    Full marks: 4/4

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    Frequently Asked Questions

    There are eight main circle theorems: the angle in a semicircle is 90 degrees; the angle at the centre is twice the angle at the circumference; angles in the same segment are equal; opposite angles of a cyclic quadrilateral add up to 180 degrees; a tangent meets a radius at 90 degrees; the two tangents from a point are equal in length; the perpendicular from the centre bisects a chord; and the alternate segment theorem. The same eight appear on the IGCSE: on Cambridge 0580 as spec points E4.7 and E4.8, and on Edexcel International GCSE 4MA1 as section 4.6.

    Eight main circle theorems, on both GCSE and IGCSE. Edexcel International GCSE 4MA1 Higher adds one idea that no other board asks for: the internal and external intersecting chord properties. Most exam questions combine two or three theorems, so it helps to know them all and to recognise which one applies from the diagram.

    The alternate segment theorem says that the angle between a tangent and a chord equals the angle in the alternate segment, which is the inscribed angle on the other side of the chord. It is often the hardest circle theorem to spot, and it appears frequently in Grade 8 and Grade 9 questions.

    On the UK GCSE they are Higher tier only, for AQA, Edexcel and OCR. On the IGCSE it is not that simple. Cambridge 0580 Core carries two of them at spec point C4.7, the angle in a semicircle and the tangent-radius right angle, and puts the rest on Extended at E4.7 and E4.8. Edexcel International GCSE 4MA1 Foundation carries three chord and tangent properties at section 4.6, and puts the angle theorems on Higher. The angle theorems, the cyclic quadrilateral and the alternate segment theorem are top-tier content on every board.

    Yes, on both major IGCSE specifications. On Cambridge IGCSE 0580 they are on the Extended tier as spec points E4.7 and E4.8, covering all eight, and on the Core tier as C4.7, which is limited to two: the angle in a semicircle is 90 degrees, and the angle between a tangent and a radius is 90 degrees. On Edexcel International GCSE Maths A (4MA1) they are section 4.6, split between Foundation, which covers the chord and tangent properties, and Higher, which adds the angle theorems, the cyclic quadrilateral, the alternate segment theorem and the intersecting chord properties.

    Look at the diagram, work out which theorem each given angle points to, write down every angle you find with its reason, and chain the steps until you reach the required angle. In the exam you must state the theorem at each step, because the reasons carry marks. Cambridge and Edexcel both say so explicitly in their specifications.

    Yes. They appear on almost every Higher tier GCSE Maths paper, and on Cambridge IGCSE 0580 Extended and Edexcel International GCSE 4MA1 Higher, usually as a multi-step find the angle giving reasons question or a short proof. They are a reliable source of marks once you know the eight theorems.

    Keep revising

    Once you are confident with circle theorems, browse the full topic lists for GCSE Maths topics and IGCSE Maths topics, and check the GCSE grade boundaries and IGCSE grade boundaries to set your target. Every topic with worked solutions is listed on the GCSE and IGCSE Maths practice questions hub. For more exam-style practice, try the Venn Diagrams Practice Questions and the Simultaneous Equations Practice Questions.

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