Circle theorems – GCSE Maths exam questions with mark scheme

1Grade 4–5HigherNon-calculator2 marks

A, B and C are points on a circle, centre O.
Angle AOB $= 124°$.

(a) Write down the size of the angle marked $x$. [1]

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

(a) $x = 62°$B1
(b) the angle at the centre is twice the angle at the circumference (subtended by the same arc)C1
Answer: $x = 62°$ — the angle at the centre is twice the angle at the circumference
Examiner tip: Learn the theorem names word for word — 'angle at centre = 2 × angle at circumference' is accepted; 'it's double' is not.

2Grade 4–5HigherNon-calculator3 marks

A, B and C are points on a circle, centre O. AB is a diameter of the circle.
Angle CAB $= 34°$.

Work out the size of the angle marked $y$.
Give reasons for your answer. [3]

angle ACB $= 90°$ (may be seen on the diagram or in working) or, on an alternative route, angle BOC $= 68°$ (or angle AOC $= 112°$ with angle OCA $= 34°$)B1
$y = 56°$ (from $180 - 90 - 34$; or via isosceles triangle OBC: angle BOC $= 68°$, $(180-68)\div 2$)B1
reasons: the angle in a semicircle is $90°$ and angles in a triangle add up to $180°$ (accept 'angle subtended by a diameter is a right angle'). Alternative routes need a complete set of correct reasons for that route, e.g. angle at the centre is twice the angle at the circumference, base angles of an isosceles triangle are equal (OB = OC, radii) and angles in a triangle add up to $180°$; or base angles of an isosceles triangle (OA = OC, radii), angles in a triangle add up to $180°$ and angles on a straight line add up to $180°$C1
Answer: $y = 56°$
Examiner tip: Spot the diameter first — a triangle drawn on a diameter always has a right angle at the circumference.

3Grade 6–7HigherNon-calculator5 marks

A, B, C and D are points on a circle. BCE is a straight line.
Angle BAD $= (2x + 10)°$ and angle BCD $= (3x - 5)°$.

(a) Work out the value of $x$.
Give a reason for your answer. [3]

(b) Work out the size of angle DCE, marked $y$.
Give a reason for your answer. [2]

(a) $2x + 10 + 3x - 5 = 180$ (or $5x + 5 = 180$)M1
(a) $x = 35$A1
(a) reason: opposite angles of a cyclic quadrilateral add up to $180°$ (do not accept 'opposite angles in a quadrilateral add up to 180')C1
(b) $y = 80°$ (angle BCD $= 3(35) - 5 = 100°$, $180 - 100$; ft $175 - 3\times$their $x$)B1
(b) reason: angles on a straight line add up to $180°$ (accept: the exterior angle of a cyclic quadrilateral equals the interior opposite angle)C1
Answer: (a) $x = 35$ (b) $y = 80°$
Examiner tip: The opposite angles of a cyclic quadrilateral add up to 180° — they are not equal. Check your answer: 80 + 100 = 180.

4Grade 6–7HigherNon-calculator5 marks

A and B are points on a circle, centre O.
TA and TB are tangents to the circle. Angle ATB $= 48°$.

Work out the size of angle OAB.
Give reasons for each stage of your working. [5]

angle OAT (or OBT) $= 90°$B1
angle AOB $= 360 - 90 - 90 - 48 \;(=132)$, or angle TAB $= (180 - 48) \div 2 \;(= 66)$M1
$(180 - 132) \div 2$, or $90 - 66$M1
angle OAB $= 24°$A1
reasons: tangent is perpendicular to the radius; angles in a quadrilateral add up to $360°$ (or tangents from an external point are equal in length); base angles of an isosceles triangle are equal (OA = OB, radii)C1
Answer: $24°$
Examiner tip: Two tangents from the same point are equal in length, so triangle TAB is isosceles — that gives a quick route to 66° and then 24°.

5Grade 8–9HigherNon-calculator5 marks

A, B and C are points on a circle. PAQ is the tangent to the circle at A.
BA = BC and angle QAC $= 64°$.

Work out the size of angle PAB, marked $z$.
Give reasons for each stage of your working. [5]

angle ABC $= 64°$B1
angle BAC $= (180 - 64) \div 2 \;(= 58)$M1
$z = 180 - 64 - 58$ (angles on a straight line), or $z$ = angle BCA by the alternate segment theoremM1
$z = 58°$A1
all reasons given: alternate segment theorem; base angles of an isosceles triangle are equal; angles in a triangle add up to $180°$; angles on a straight line add up to $180°$C1
Answer: $z = 58°$
Examiner tip: The alternate segment theorem links the tangent–chord angle to the angle in the OTHER segment — find the chord first, then look across it.

6Grade 8–9HigherNon-calculator4 marks

A, B and C are points on a circle, centre O. The line CO is extended to D.

Prove that angle AOB $= 2 \times$ angle ACB.
Give a reason for each stage of your proof. [4]

let angle OCA $= a$; OA = OC (radii) so triangle OAC is isosceles and angle OAC $= a$ (base angles of an isosceles triangle are equal)M1
angle AOD $= 2a$ (exterior angle of a triangle equals the sum of the interior opposite angles); or angle AOC $= 180 - 2a$ (angles in a triangle) and then angle AOD $= 2a$ (angles on a straight line)M1
similarly, with angle OCB $= b$: angle OBC $= b$ and angle BOD $= 2b$ (or angle BOC $= 180 - 2b$)M1
complete proof: angle ACB $= a + b$, so angle AOB $= 2a + 2b = 2(a + b) = 2 \times$ angle ACB (or AOB $= 360 - (180 - 2a) - (180 - 2b) = 2a + 2b$, angles around a point add up to $360°$), with a correct reason given for every stage. The M marks may be awarded for correct method without reasons; this mark needs all the reasonsC1
Answer: Using isosceles triangles OAC and OBC: AOD $= 2a$, BOD $= 2b$, so AOB $= 2(a+b) = 2\,$ACB
Examiner tip: In a proof, every angle fact needs a reason — 'OA = OC because they are radii' is the step most often left out.

7Grade 6–7HigherNon-calculator4 marks

A, B, C and D are points on a circle. The chords AC and BD meet at X.
Angle ADB $= 41°$ and angle DBC $= 33°$.

Work out the size of angle BXC, marked $w$.
Give reasons for your answer. [4]

angle ACB $= 41°$ (or angle DAC $= 33°$)B1
$180 - 41 - 33$ (in triangle BXC, or in triangle AXD to find angle AXD)M1
$w = 106°$A1
reasons: angles in the same segment are equal (accept 'angles subtended by the same arc are equal') and angles in a triangle add up to $180°$ (and vertically opposite angles are equal, if the AXD route is used)C1
Answer: $w = 106°$
Examiner tip: Look for a 'bow-tie': angles standing on the same chord, on the same side, are equal. Say 'angles in the same segment', not just 'same arc angles'.

8Grade 1–3FoundationNon-calculator3 marks

The diagram shows a circle, centre O.
A, B and T are points on the circle.

(a) Write down the mathematical name of the straight line AB. [1]

(b) The straight line PQ touches the circle at T only.
Write down the mathematical name of the line PQ. [1]

(a) chordB1
(b) tangentB1
(c) segment (accept minor segment)
Do not accept sectorB1
Answer: (a) chord (b) tangent (c) segment
Examiner tip: A segment is cut off by a chord; a sector is cut off by two radii — students mix them up every year.

9Grade 1–3Non-calculator3 marks

A circle has a radius of $6.5$ cm.

(a) Write down the length of the longest chord that can be drawn in this circle. [1]

(b) Lena says,
“A sector and a segment are the same thing. They are both parts of a circle cut off by a straight line.”

Explain why Lena is wrong. [1]

(a) $13$ (cm)B1
(b) explanation that a sector is bounded by two radii and an arc, whereas a segment is bounded by a chord and an arc
Accept a correct description of either one, e.g. “a sector is between two radii” or “a sector is a pizza-slice shape that reaches the centre, a segment does not”C1
(c) arcB1
Answer: (a) $13$ cm (b) a sector is bounded by two radii and an arc; a segment by a chord and an arc (c) arc
Examiner tip: The longest chord is always the diameter, so double the radius — do not write $6.5$.

10Grade 6–7HigherCalculator5 marks

T is a point on a circle, centre O.
PT is the tangent to the circle at T.
OT $= 8.4$ cm and PT $= 15.2$ cm.

(a) Work out the length of OP.
Give your answer correct to 1 decimal place.
You must give a reason for any angle facts you use. [3]

(b) Work out the size of angle TOP.
Give your answer correct to 1 decimal place. [2]

(a) angle OTP $= 90°$ with reason: a tangent is perpendicular to the radius (at the point of contact)B1
$8.4^2 + 15.2^2 \;(= 301.6)$ or $\sqrt{8.4^2 + 15.2^2}$M1
$17.4$ (cm) (accept $17.36$ to $17.37$ if not rounded)A1
(b) $\tan(\text{TOP}) = \dfrac{15.2}{8.4}$ or $\cos(\text{TOP}) = \dfrac{8.4}{\text{their OP}}$ or $\sin(\text{TOP}) = \dfrac{15.2}{\text{their OP}}$M1
$61.1°$ (accept $60.9°$ to $61.1°$ from correct use of a rounded OP)A1
Answer: (a) $17.4$ cm (b) $61.1°$
Examiner tip: You can only use Pythagoras because the tangent meets the radius at $90°$ — say so, or the reason mark is lost.

11Grade 6–7HigherNon-calculator5 marks

A, B, C and D are points on a circle, centre O.
Angle ACB $= (2x + 14)°$ and angle ADB $= (5x - 31)°$.

(a) Work out the value of $x$.
Give a reason for your answer. [3]

(b) Work out the size of angle AOB.
Give a reason for your answer. [2]

(a) $2x + 14 = 5x - 31$M1
$x = 15$A1
reason: angles in the same segment are equal (accept “angles subtended by the same arc (or chord) are equal”; do not accept “angles at the circumference are equal” alone)C1
(b) angle AOB $= 88°$ (ft $2 \times (2 \times \text{their } x + 14)$)B1
reason: the angle at the centre is twice the angle at the circumference (subtended by the same arc)C1
Answer: (a) $x = 15$ (b) $88°$
Examiner tip: The answer to (a) is $x$, not the angle — but you need the angle ($44°$) before you can double it in (b).

12Grade 4–5HigherNon-calculator4 marks

A, B, C and D are points on a circle, centre O.
Angle ADC $= 112°$.

Sam writes,
“Angle ABC $= 112°$ because opposite angles in a cyclic quadrilateral are equal.”

(a) Explain what is wrong with Sam’s statement. [1]

(b) Work out the correct size of angle ABC. [1]

(a) opposite angles in a cyclic quadrilateral add up to $180°$ (they are not equal)C1
(b) $68°$B1
(c) $136°$ (ft $2 \times$ their (b) provided it is less than $180$)B1
reason: the angle at the centre is twice the angle at the circumferenceC1
Answer: (a) opposite angles of a cyclic quadrilateral sum to $180°$, not equal (b) $68°$ (c) $136°$
Examiner tip: Use angle ABC (not ADC) for the centre angle here: doubling $112°$ gives the reflex angle AOC, $224°$.

13Grade 8–9HigherNon-calculator4 marks

A, B, C and D are points on a circle, centre O.
ABCD is a cyclic quadrilateral.
Angle ADC $= x$ and angle ABC $= y$.

Prove that $x + y = 180°$.
Do not use the fact that opposite angles of a cyclic quadrilateral add up to $180°$.
Give a reason for each stage of your proof. [4]

angle AOC (the angle at O on the side of B, subtended by arc ABC) $= 2x$M1
reflex angle AOC $= 360° - 2x$M1
$y = \dfrac{360° - 2x}{2} = 180° - x$ (angle at the centre applied to arc ADC)
Allow labelling the two angles at O as $2x$ and $2y$ and using $2x + 2y = 360°$M1
complete proof concluding $x + y = 180°$ with no incorrect statements and reasons given: the angle at the centre is twice the angle at the circumference; angles around a point add up to $360°$C1
Answer: Angle AOC $= 2x$ (centre angle); reflex AOC $= 360° - 2x$ (angles at a point); $y = \tfrac{1}{2}(360° - 2x) = 180° - x$, so $x + y = 180°$
Examiner tip: You may not quote the result you are proving — build it from the angle-at-the-centre theorem applied twice.

14Grade 4–5HigherCalculator6 marks

A and B are points on a circle, centre O. The radius of the circle is $11$ cm.
AB is a chord of length $18$ cm.
M is the point on AB such that OM is perpendicular to AB.

(a) Write down the length of AM.
Give a reason for your answer. [2]

(b) Work out the length of OM.
Give your answer correct to 3 significant figures. [2]

(a) AM $= 9$ (cm)B1
reason: the perpendicular from the centre to a chord bisects the chord (accept “OAB is isosceles because OA = OB are radii, so OM bisects AB”)C1
(b) $11^2 - 9^2 \;(= 40)$ or $\sqrt{11^2 - 9^2}$M1
$6.32$ (cm)A1
(c) $\sin(\text{AOM}) = \dfrac{9}{11}$ (or $\cos(\text{AOM}) = \dfrac{\text{their OM}}{11}$ or $\tan(\text{AOM}) = \dfrac{9}{\text{their OM}}$) and doubling, or $\cos(\text{AOB}) = \dfrac{11^2 + 11^2 - 18^2}{2 \times 11 \times 11}$M1
$109.8°$ (accept $109.7°$ to $109.9°$)A1
Answer: (a) $9$ cm (perpendicular from the centre bisects the chord) (b) $6.32$ cm (c) $109.8°$
Examiner tip: In (c) the trig in triangle AOM gives only half of angle AOB — remember to double it.

15Grade 8–9HigherNon-calculator5 marks

A, B and C are points on a circle, centre O.
AB is a diameter of the circle.
The tangent to the circle at C meets the line AB extended at T.
Angle CTA $= 34°$.

Work out the size of angle CAB.
Give a reason for each stage of your working. [5]

angle OCT $= 90°$
Alternative route: angle ACB $= 90°$B1
angle COT $= 180 - 90 - 34 \;(= 56)$, or angle AOC $= 180 - 56 \;(= 124)$
Alternative route: angle BCT $=$ angle CAB (alternate segment theorem), e.g. both labelled $a$M1
$(180 - 124) \div 2$ or $56 \div 2$
Alternative route: $a + 34 + 90 + a = 180$ in triangle ACT (so $2a = 56$)M1
angle CAB $= 28°$A1
reasons for their route, e.g. a tangent is perpendicular to the radius; angles in a triangle add up to $180°$; angles on a straight line add up to $180°$; base angles of an isosceles triangle are equal (OA = OC are radii) or the exterior angle of a triangle equals the sum of the interior opposite angles or the angle at the centre is twice the angle at the circumference
Alternative route: angle in a semicircle is $90°$; alternate segment theorem; angles in a triangle add up to $180°$C1
Answer: $28°$
Examiner tip: Use the radius OC: it meets the tangent at $90°$ and makes triangle AOC isosceles — and give a reason at every step.

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