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Circle theorems

Free Printable Circle theorems PDF Worksheets, Revision Guide & GCSE Past Paper Questions (Grade 6)

Download free circle theorems worksheets (PDF), GCSE maths past paper questions and worked solutions for Circle theorems (6K). This is a Grade 6 geometry topic from the GCSE questions by topic collection for 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, Maths4Everyone, Corbett Maths, Maths Made Easy, Dr Frost, Naiker Maths, 1st Class Maths, GCSE Maths Rocket, Dr Austin, Metatutor, E Classroom and PMT, along with past paper question booklets.

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

Free Circle theorems PDF Worksheets & Answers

Grade 6 • Topic 6K
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Grade 6 topics
6A - Recurring decimals6B - Harder indices
6C - The product rule6D - Repeated change
6E - Triple brackets6F - Parallel and perpendicular
6G - Inequalities on graphs6H - Similar shapes (area and vol)
6J - Negative scale factors6K - Circle theorems
6L - Cumulative frequency6M - Box plots
6N - Combined transformations6P - Capture recapture
GCSE Maths Grade 6 Revision Notes & Method

Guide to Circle theorems: Explanation, Diagrams & Worked Examples

Download Printable One Page Guide (PDF)

Explanation:

Key circle theorems (always write the exact worded reason in exams): • Angle at the centre is twice the angle at the circumference. • Angle in a semicircle is 90°. • Angles in the same segment are equal. • Opposite angles in a cyclic quadrilateral add up to 180°.

For tangents: a tangent meets a radius at 90°; tangents from an external point are equal in length; and by the alternate segment theorem, the angle between a tangent and a chord equals the angle subtended by the chord in the alternate segment.

Example 1: A, B, C, D lie on a circle with centre O. Angle AOC = 136°. Find acute angle ABC and obtuse angle ADC.

O 136° 68° A B C D (112°)

Solution:

∠ABC = 136° ÷ 2 = 68° (angle at centre = 2 × circumference)
∠ADC = 180° − 68° = 112° (opposite angles in cyclic quad = 180°)
Answer: ∠ABC = 68°, ∠ADC = 112°

Example 2: PA and PB are tangents from P to a circle with centre O. Given ∠APB = 50°, find ∠OAB.

O A B P 50°

Solution:

PA = PB (▵PAB isosceles) → ∠PAB = (180° − 50°) ÷ 2 = 65°
∠OAP = 90° (tangent ⊥ radius) → ∠OAB = 90° − 65° = 25°
Answer: ∠OAB = 25°
Scanned GCSE Exam Questions & Worked Solutions

GCSE Past Paper Circle theorems Questions & Answers

Past Paper Example 1 GCSE Nov 2021 2H Q14 4 marks
Full Paper (PDF) • Mark Scheme (PDF)
20°SBTADC
A, B, C and D are four points on a circle.
SBT is a tangent to the circle.
Angle ABD = 20°
the size of angle BAD : the size of angle BCD = 3 : 1
Find the size of angle SBA.
Give a reason for each stage of your working.
A, B, C and D are four points on a circle.
SBT is a tangent to the circle.
Angle ABD = 20°
the size of angle BAD : the size of angle BCD = 3 : 1
Find the size of angle SBA.
Give a reason for each stage of your working.
Worked Solution & Mark Scheme (4 marks)
Opposite angles of a cyclic quadrilateral add up to 180°, so angle BCD = 180° ÷ 4 = 45° and angle BAD = 135° M2
Angle SBD = angle BCD = 45° (alternate segment theorem) M1
Therefore angle SBA = 45° – 20° = 25°. A1
Answer: 25
Past Paper Example 2 GCSE Nov 2020 3H Q14 4 marks
Full Paper (PDF) • Mark Scheme (PDF)
ABCDEF42°104°
Points B, D, E and F lie on a circle.
ABC is the tangent to the circle at B.
Find the size of angle ABD.
You must give a reason for each stage of your working.
Points B, D, E and F lie on a circle.
ABC is the tangent to the circle at B.
Find the size of angle ABD.
You must give a reason for each stage of your working.
Worked Solution & Mark Scheme (4 marks)
Angle DBF = 180° – 104° = 76° (opposite angles of a cyclic quadrilateral add up to 180°) M2
Angle BFD = 180° – (76° + 42°) = 62° (angles in a triangle add up to 180°) M1
Angle ABD = 62° (alternate segment theorem) C1
Answer: 62° with full geometric reasons
Past Paper Example 3 AQA Jun 2023 2H Q20 3 marks
Full Paper (PDF) • Mark Scheme (PDF)
(a) Assume S is the centre and angle PSR = 130°. Work out x. [1 mark]
20 (b) Centre is on PS but not at S. What does this mean about x? [1 mark]
20 (c) Simon says Angle ADC = 70° (alternate segment theorem) so y = 70°. Is he correct? [1 mark]
(a) P, Q and R are points on a circle.
S is a point inside triangle PQR. [1 mark]
xS130°QPRNot drawnaccurately
Assume that S is the centre of the circle. Work out the size of angle x.
(b) In fact, the centre of the circle is on PS but **not** at S.
What does this mean about the size of angle x ? Tick **one** box.
[ ] It is the same as the answer to part (a)
[ ] It is greater than the answer to part (a)
[ ] It is smaller than the answer to part (a)
[ ] It is impossible to tell [1 mark]
(c) For a different circle,
• AB is a tangent at A
• C and D are on the circumference of the circle
• AC = CD [1 mark]
y70°ABCDNot drawnaccurately
Here is Simon’s method to work out the size of angle y.
@@TABULAR0@@
Is he correct? Give a reason for your answer.
Worked Solution & Mark Scheme (3 marks)
Part (a) Solution:
Angle at circumference = 130° ÷ 2 = 65° B1
Answer (a): x = 65
Part (b) Solution:
The centre could be between P and S or beyond S, so it is impossible to tell B1
Answer (b): It is impossible to tell
Part (c) Solution:
No, the angle in the alternate segment to 70° is angle ACD = 70° (so y = 55°) C1
Answer (c): No — angle ACD = 70° by the alternate segment theorem (not angle ADC), so y = (180° – 70°) ÷ 2 = 55°

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