Circle theorems Quiz (Grade 6)

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Score: 0 / 10

Circle theorems — Notes & Worked Examples Show / Hide

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.

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.

Solution:

PA = PB (▵PAB isosceles) → ∠PAB = (180° − 50°) ÷ 2 = 65°

∠OAP = 90° (tangent ⊥ radius) → ∠OAB = 90° − 65° = 25°

Answer: ∠OAB = 25°

One Page Guide (PDF) Circle theorems Worksheets

Question 1

[2 marks]

A, B and C are points on the circumference of a circle with centre O.
Angle ACB = 40°. Work out the size of angle AOB.

Worked Solution & Answer

(The circumference of a circle is π × diameter = 2πr)
The angle at the centre is twice the angle at the circumference: 2 × 40° = 80°. B1

Answer: 80° °

Question 2

[2 marks]

A, B and C are points on the circumference of a circle with centre O.
Angle ACB = 64°. Work out the size of angle AOB.

Worked Solution & Answer

(The circumference of a circle is π × diameter = 2πr)
The angle at the centre is twice the angle at the circumference: 2 × 64° = 128°. B1

Answer: 128° °

Question 3 IGCSE Nov 2021 2H Q11

[3 marks]

B, D, E and F are points on a circle, centre O.
ABC is a tangent to the circle.
ODC is a straight line.
BOE is a diameter of the circle.
Angle BCD = 48°
Find the size of angle DFE.

Worked Solution & Answer

(Angles on a straight line add up to 180°)
∠OBC = 90° (tangent-radius) ⇒ ∠BOD = 180 – 90 – 48 = 42° M1
Since BOE is a diameter, ∠DOE = 180 – 42 = 138° M1
Angle at circumference ∠DFE = 138 ÷ 2 = 69° A1

Answer: 69 °

Question Paper (PDF) • Mark Scheme (PDF)

Question 4 AQA Jun 2023 2H Q20

[3 marks]

(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]

Assume that S is the centre of the circle. Work out the size of angle x.

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]

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 & Answer

(a)

Angle at circumference = 130° ÷ 2 = 65° B1

Answer (a): x = 65

(b)

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

(c)

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°

Question Paper (PDF) • Mark Scheme (PDF)

Question 5

[4 marks]

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 & Answer

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

Question Paper (PDF) • Mark Scheme (PDF)

Question 6

[4 marks]

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 & Answer

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

Question Paper (PDF) • Mark Scheme (PDF)

Question 7 OCR Jun 2023 4H Q17

[4 marks]

Work out
(a) (i)
(ii) Work out angle b.
Give a reason for your answer.

(a) [2 marks]

Angle PRQ = 59° and angle PQR = 67°.
Line SPT is a tangent to the circle.
Work out angle a.
Give a reason for your answer.

(b) Work out angle b.
Give a reason for your answer. [2 marks]

Worked Solution & Answer

(a)

(The circumference of a circle is π × diameter = 2πr)
Angle a = angle PQR = 67° because angles in the alternate segment are equal. M1 C1

Answer (a): Angle a = 67° because angles in the alternate segment are equal.

(b)

Angle OPT = 90° because the angle between a radius and a tangent is 90° M1
Therefore b = 90° − 67° = 23°. C1

Answer (b): Angle b = 23° because the angle between a tangent and a radius is 90° (90° − 67° = 23°).

Question Paper (PDF) • Mark Scheme (PDF)

Question 8 IGCSE Jun 2018 1H Q13

[4 marks]

L, M, N on circle centre O, tangent QMT, angle NLM = 27°. (a)(i) Angle NOM (1) (ii) Reason (1) (b)(i) Angle NMQ (1) (ii) Reason (1)

L, M and N are points on a circle, centre O.
QMT is the tangent to the circle at M.

(a) Find the size of angle NOM. [1 mark]

(b) Give a reason for your answer. [1 mark]

(b) Find the size of angle NMQ. [1 mark]

(d) Give a reason for your answer. [1 mark]

Worked Solution & Answer

(a)

2 × 27° = 54° B1

Answer (a): 54 °

(b)

The angle at the centre is twice the angle at the circumference

Answer (b): The angle at the centre is twice the angle at the circumference

(b)

Angle NMQ = angle NLM = 27° by alternate segment theorem (or ∠OMN = (180–54)/2 = 63°, ∠NMQ = 90–63 = 27°) B1

Answer (b): 27 °

(d)

Alternate segment theorem (or angle between tangent and radius is 90°)

Answer (d): Alternate segment theorem (or angle between tangent and radius is 90°)

Question Paper (PDF) • Mark Scheme (PDF)

Question 9 IGCSE Jan 2023 2H Q14

[5 marks]

A, B, C and D are points on a circle, centre O
EBF is the tangent to the circle at B

(a) Work out the size of angle DCB [1 mark]

(b) Give a reason for your answer to (a)(i) [1 mark]

(b) Work out the size of angle ADO [3 marks]

Worked Solution & Answer

(a)

180° − 40° = 140° (opposite angles of a cyclic quadrilateral sum to 180°) B1

Answer (a): 140 °

(b)

Opposite angles of a cyclic quadrilateral sum to 180°

Answer (b): Opposite angles of a cyclic quadrilateral sum to 180°

(b)

Angle DOB = 2 × 40° = 80° (angle at centre) M1
Since OD = OB, angle ODB = (180° − 80°)/2 = 50° M1
By alternate segment theorem, angle ADB = angle ABF = 66°
Therefore angle ADO = 66° − 50° = 16° A1

Answer (b): 16 °

Question Paper (PDF) • Mark Scheme (PDF)

Question 10 IGCSE Nov 2024 2H Q16

[5 marks]

A, B, C and D are points on a circle, centre O

(a) Work out the size of angle AOD [1 mark]

(b) Give a reason for your answer to part (a)(i) [1 mark]

(b) Work out the size of angle CAO [1 mark]

Worked Solution & Answer

(a)

2 × 54° = 108° B1

Answer (a): 108 °

(b)

The angle at the centre is twice the angle at the circumference

Answer (b): The angle at the centre is twice the angle at the circumference

(b)

Angle OAD = (180° − 108°)/2 = 36°
in quadrilateral AOCD or via reflex AOD = 252°, angle CAO = 54° − 28° = 26° B1

Answer (b): 26 °

(c)

Angle ADC = 28° + 36° = 64° M1
opposite angles in cyclic quadrilateral ABCD sum to 180°, so angle ABC = 180° − 64° = 116° A1

Answer (c): 116 °

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Score: 0 / 10 (0 / 36 marks)

Circle theorems Worksheets Next: Cumulative frequency →

Grade 6 Quizzes

14 Topics

6A - Recurring decimals	6B - Harder indices
6C - The product rule	6D - Repeated change
6E - Triple brackets	6F - Parallel and perpendicular
6G - Inequalities on graphs	6H - Similar shapes (area and vol)
6J - Negative scale factors	6K - Circle theorems
6L - Cumulative frequency	6M - Box plots
6N - Combined transformations	6P - Capture recapture