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Circle theorem proof

Free Printable Circle theorem proof PDF Worksheets, Revision Guide & GCSE Past Paper Questions (Grade 8+)

Download free circle theorem proof worksheets (PDF), GCSE maths past paper questions and worked solutions for Circle theorem proof (8J). This is a Grade 8+ geometry topic from the GCSE questions by topic collection for Foundation and Higher tier GCSE Maths (Edexcel, AQA, OCR and IGCSE).

The free PDF worksheets table below brings together printable question booklets (Qs) and worked answers or mark schemes (Ans) from Maths Genie, Corbett Maths, Naiker Maths, 1st Class Maths, E Classroom and PMT, along with past paper question booklets.

Below the worksheet links you can read our circle theorem proof revision notes and worked examples with diagrams, try real GCSE past paper circle theorem proof questions with step-by-step solutions, or download a PDF of all circle theorem proof worksheet links.

Free Circle theorem proof PDF Worksheets & Answers

Grade 8+ • Topic 8J
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Exam question booklet & solutions
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Edexcel exam questions & solutions
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PMT (Edxl)
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AQA GCSE topic questions & solutions
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Grade 8+ topics
8A - Quad. sim. equations8B - Transforming graphs y=f(x)
8C - Algebraic proof8D - Completing the square
8E - Quadratic inequalities8F - Quadratic sequences
8G - Velocity time graphs8H - Circles and tangents
8J - Circle theorem proof8K - Vector proof
8L - Probability equation questions8M - Invariant points
8N - Spheres, cones and cylinders8P - Volume of a frustum
GCSE Maths Grade 8+ Revision Notes & Method

Guide to Circle theorem proof: Explanation, Diagrams & Worked Examples

Download Printable One Page Guide (PDF)

Explanation:

Higher tier papers can ask you to prove standard circle theorems from first principles. The key construction is to draw radii from the centre O to the vertices on the circumference, creating isosceles triangles (since all radii are equal).

Label the equal base angles of the isosceles triangles as x and y, then use angles in a triangle (180°) and angles around a point (360°) or tangent-radius (90°) to link the angles algebraically. Always state geometric reasons at every step.

Example 1: Prove that opposite angles in a cyclic quadrilateral ABCD add up to 180°.

A x B D C y 2x 2y

Solution:

Join B, D to O. Let ∠BAD = x, ∠BCD = y ⇒ ∠BOD = 2x and 2y
Angles around O: 2x + 2y = 360° ⇒ x + y = 180°

Example 2: Prove the alternate segment theorem: for tangent PT at A and chord AB, ∠BAT = ∠ACB.

O 2x C x B A P T x

Solution:

Let ∠BAT = x. ∠OAT = 90° ⇒ ∠OAB = ∠OBA = 90° − x
∠AOB = 180° − 2(90° − x) = 2x ⇒ ∠ACB = 12(2x) = x
Scanned GCSE Exam Questions & Worked Solutions

GCSE Past Paper Circle theorem proof Questions & Answers

Past Paper Example 1 AQA Nov 2022 1H Q26 4 marks
Full Paper (PDF) • Mark Scheme (PDF)
P, Q and R are points on a circle.
SP is a tangent to the circle.
RQ = PQ
y2xSPRQNot drawnaccurately
Prove that y = 90° - x
Worked Solution & Mark Scheme (4 marks)
By the alternate segment theorem, angle PQR = angle RPS = 2x M2
Since RQ = PQ, triangle PQR is isosceles, so angle QPR = angle PRQ = y M1
Angles in triangle PQR add to 180°: y + y + 2x = 180° ⇒ 2y = 180° – 2x ⇒ y = 90° – x. C1
Answer: Angle PQR = 2x (alternate segment theorem).
Angle QPR = y (isosceles triangle PQR as RQ = PQ).
2y + 2x = 180° (angles in a triangle sum to 180°), so y = 90° – x.
Past Paper Example 2 GCSE Nov 2017 3H Q20 4 marks
Full Paper (PDF) • Mark Scheme (PDF)
ABCO
A, B and C are points on the circumference of a circle, centre O. AOB is a diameter. Prove that angle ACB is 90° without using circle theorems.
A, B and C are points on the circumference of a circle, centre O.
AOB is a diameter of the circle.
Prove that angle ACB is 90°
You must **not** use any circle theorems in your proof.
Worked Solution & Mark Scheme (4 marks)
(The circumference of a circle is π × diameter = 2πr)
Join O to C M1
OA = OC = OB (radii of circle), so triangles OAC and OBC are isosceles M1
Let ∠OAC = ∠OCA = x and ∠OBC = ∠OCB = y (base angles of isosceles triangles are equal) M1
Sum of angles in triangle ABC = x + (x + y) + y = 2x + 2y = 180° ⇒ x + y = 90°, so ∠ACB = 90°. C1
Answer: Draw radius OC. OA = OC = OB (radii), so △ OAC and △ OBC are isosceles. Let ∠ OAC = ∠ OCA = x and ∠ OBC = ∠ OCB = y (base angles of an isosceles triangle are equal). In △ ABC, x + (x + y) + y = 180° (angles in a triangle sum to 180°) ⇒ 2(x + y) = 180° ⇒ ∠ ACB = x + y = 90°.
Past Paper Example 3 GCSE Jun 2024 2H Q22 4 marks
Full Paper (PDF) • Mark Scheme (PDF)
A and B are points on a circle, centre O.
MAP and NBP are tangents to the circle.
OMNABP
Prove that AP = BP
Worked Solution & Mark Scheme (4 marks)
Join OA, OB and OP. ∠OAP = ∠OBP = 90° (angle between radius and tangent is 90°) M1
OA = OB (both radii) M1
OP is a common side M1
By RHS, ΔOAP ≅ ΔOBP, hence AP = BP. C1
Answer: In triangles OAP and OBP: ∠OAP = ∠OBP = 90° (angle between tangent and radius is 90°), OA = OB (radii of the same circle), and OP is common. Therefore triangles OAP and OBP are congruent by RHS, so AP = BP.

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