Circle properties & circle theorems with reasons – IGCSE Maths exam questions with mark scheme

1Grade 1–3FoundationCalculator4 marks

The diagram shows a circle with centre $O$.
$A$, $B$, $C$, $D$ and $T$ are points on the circle.
The straight line $L$ touches the circle at $T$.

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

(b) Write down the mathematical name of the line $L$. [1]

(d) Write down the mathematical name of the region $S$, which is bounded by $OC$, $OD$ and the arc $CD$. [1]

(a) chord (do not accept diameter or line segment alone)B1
(b) tangentB1
(c) segment (accept minor segment; do not accept sector or arc)B1
(d) sector (accept minor sector; do not accept segment)B1
Answer: (a) chord (b) tangent (c) segment (d) sector
Examiner tip: A segment is cut off by a chord, a sector is cut off by two radii — swapping these two words is the commonest lost mark.

2Grade 4–5Calculator6 marks

$A$ and $B$ are points on a circle, centre $O$.
$PA$ and $PB$ are tangents to the circle.
Angle $APB = 46^\circ$.

(a) Work out the size of angle $PAB$.
Give a reason for your answer. [3]

(b) Work out the size of angle $AOB$.
Give reasons for your working. [3]

(a) $(180 - 46) \div 2$ oeM1
(a) $67^\circ$A1
(a) reason: tangents from a point (outside a circle) are equal in length, so triangle $PAB$ is isosceles (accept 'base angles of an isosceles triangle are equal' only if $PA = PB$ is justified by tangents)B1
(b) $360 - 90 - 90 - 46$ oe (may be seen as angles marked $90^\circ$ on the diagram)M1
(b) $134^\circ$A1
(b) reasons: a tangent is perpendicular to the radius (at the point of contact) and angles in a quadrilateral add up to $360^\circ$B1
Answer: (a) $67^\circ$ (b) $134^\circ$
Examiner tip: Write the full reason 'tangent meets radius at $90^\circ$' — simply marking right angles on the diagram earns the method mark but not the reason mark.

3Grade 4–5Calculator4 marks

$A$ and $B$ are points on a circle, centre $O$.
$AB$ is a chord of the circle.
$M$ is the point on $AB$ such that $OM$ is perpendicular to $AB$.
$AB = 19.6$ cm and $OM = 5.2$ cm.

(a) Give a reason why $AM = 9.8$ cm. [1]

(b) Work out the radius of the circle.
Give your answer correct to 3 significant figures. [3]

(a) the perpendicular from the centre (of a circle) to a chord bisects the chord (accept '$M$ is the midpoint of $AB$ because $OM$ is perpendicular to the chord from the centre'; do not accept just 'half of 19.6')B1
(b) $9.8^2 + 5.2^2$ oe ($= 123.08$)M1
(b) $\sqrt{9.8^2 + 5.2^2}$ oeM1
(b) $11.1$ cm (accept answers rounding to $11.09$)A1
Answer: (b) $11.1$ cm
Examiner tip: The radius is the hypotenuse $OA$ of triangle $OMA$, so use half the chord ($9.8$), not the full $19.6$.

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4Grade 4–5HigherCalculator4 marks

$A$, $B$ and $C$ are points on a circle, centre $O$.
$AOC$ is a diameter of the circle.
Angle $BAC = 33^\circ$.

(a) (i) Work out the size of angle $ACB$.
(ii) Give reasons for your answer. [2]

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

(a)(i) $57^\circ$B1
(a)(ii) angle in a semicircle is $90^\circ$ (accept 'angle $ABC = 90^\circ$ because $AC$ is a diameter') and angles in a triangle add up to $180^\circ$B1
(b)(i) $114^\circ$ (ft $2 \times$ their (a)(i))B1
(b)(ii) the angle at the centre is twice the angle at the circumference (subtended by the same arc) — accept alternative: $OA = OB$ (radii) so base angles of an isosceles triangle are equal, and angles in a triangle add up to $180^\circ$B1
Answer: (a) $57^\circ$ (b) $114^\circ$
Examiner tip: Name the theorem in full: 'angle in a semicircle is $90^\circ$', not just 'it's a diameter'.

5Grade 6–7HigherCalculator5 marks

$A$, $B$, $C$ and $D$ are points on a circle.
Angle $ABC = (4x - 7)^\circ$
Angle $ADC = (2x + 25)^\circ$
Angle $BAC = 41^\circ$

(a) Work out the value of $x$. [2]

(b) Angle $ADB = y^\circ$.
Work out the value of $y$.
Give reasons for each stage of your working. [3]

(a) $4x - 7 + 2x + 25 = 180$ oeM1
(a) $x = 27$A1
(b) angle $ABC = 101^\circ$ (ft their $x$) and $180 - 101 - 41$ oe for angle $ACB$M1
(b) $y = 38$A1
(b) reasons: angles in a triangle add up to $180^\circ$ and angles in the same segment are equal (accept 'angles subtended by the same arc are equal'; do not accept 'angles from the same chord' without 'segment' or 'arc')B1
Answer: (a) $x = 27$ (b) $y = 38$
Examiner tip: Angle $ADB$ and angle $ACB$ stand on the same arc $AB$ — the reason must say 'angles in the same segment are equal'.

6Grade 6–7HigherCalculator4 marks

$A$, $B$ and $C$ are points on a circle.
$SAT$ is the tangent to the circle at $A$.
Angle $TAC = 52^\circ$ and angle $CAB = 64^\circ$.

Show that triangle $ABC$ is isosceles.
Give reasons for each stage of your working. [4]

angle $ABC = 52^\circ$ (may be marked on the diagram) — or angle $SAB = 180 - 52 - 64 = 64^\circ$B1
angle $ACB = 180 - 52 - 64$ oe — or angle $ACB = $ angle $SAB$M1
angle $ACB = 64^\circ$ = angle $CAB$, so $BA = BC$ and triangle $ABC$ is isosceles (conclusion must be stated)A1
reasons: alternate segment theorem and angles in a triangle add up to $180^\circ$ (or angles on a straight line add up to $180^\circ$ for the $SAB$ route)B1
Answer: Angle $ABC = 52^\circ$ (alternate segment theorem), so angle $ACB = 64^\circ$ = angle $CAB$; two equal angles, so triangle $ABC$ is isosceles.
Examiner tip: In a 'show that', finding $64^\circ$ is not enough: you must say that two angles are equal and conclude that the triangle is isosceles.

7Grade 6–7HigherCalculator4 marks

$A$, $B$, $C$ and $D$ are points on a circle.
The chords $AB$ and $CD$ intersect at the point $P$.
$AP = x$ cm, $PB = (x + 5)$ cm, $CP = 3$ cm and $PD = 12$ cm.

Work out the length of $AB$.
Show clear algebraic working. [4]

$x(x + 5) = 3 \times 12$ oe (intersecting chords: $AP \times PB = CP \times PD$)M1
$x^2 + 5x - 36 = 0$ and a correct method to solve, e.g. $(x + 9)(x - 4) = 0$M1
$x = 4$ (ignore $x = -9$ if rejected or not used)A1
$AB = 13$ cm (ft $2x + 5$ for their positive $x$)B1
Answer: $AB = 13$ cm
Examiner tip: Many stop at $x = 4$ — the question asks for $AB$, which is $x + (x + 5) = 13$ cm.

8Grade 8–9HigherCalculator4 marks

$A$, $B$, $C$ and $D$ are points on a circle.
$PAB$ and $PCD$ are straight lines.
$PA = x$ cm, $AB = 9$ cm, $PC = 6$ cm and $CD = 4$ cm.

Work out the value of $x$.
Give your answer correct to 3 significant figures. [4]

$x(x + 9) = 6 \times 10$ oe (i.e. $PA \times PB = PC \times PD$)M1
$x^2 + 9x - 60 = 0$ (correct 3-term quadratic, may be rearranged)M1
correct method to solve their quadratic, e.g. $\dfrac{-9 \pm \sqrt{9^2 - 4 \times 1 \times (-60)}}{2}$ or $\left(x + 4.5\right)^2 = 80.25$M1
$4.46$ (accept answers rounding to $4.458$; ignore the negative root if rejected)A1
Answer: $x = 4.46$
Examiner tip: Outside the circle you multiply the whole line lengths: $PA \times PB$ uses $PB = x + 9$, not $AB = 9$ — writing $9x = 6 \times 4$ scores nothing.

9Grade 8–9HigherCalculator5 marks

$A$, $B$ and $C$ are points on a circle, centre $O$.
$TA$ is the tangent to the circle at $A$.
Angle $TAC = 54^\circ$ and angle $ACB = 79^\circ$.

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

finds one relevant angle, e.g. angle $ABC = 54^\circ$ or angle $OAC = 36^\circ$ or angle $AOC = 108^\circ$ (may be on the diagram)M1
complete method, e.g. angle $BAC = 180 - 54 - 79 = 47$, angle $BOC = 94$, then $(180 - 94) \div 2$; or angle $OCA = 36$, then $79 - 36 = 43$ and angle $OBC$ = angle $OCB$M1
$43^\circ$A1
at least one correct reason relevant to their method, e.g. alternate segment theorem; tangent is perpendicular to the radiusB1
a full set of reasons for their method, e.g. alternate segment theorem, angles in a triangle add up to $180^\circ$, angle at centre is twice the angle at the circumference, base angles of an isosceles triangle are equal ($OB = OC$, radii)B1
Answer: $43^\circ$
Examiner tip: Reasons only score if they match the step they justify — 'isosceles triangle' needs the note that $OB = OC$ because both are radii.

10Grade 4–5HigherCalculator3 marks

$P$, $Q$, $R$ and $S$ are points on a circle.
Angle $QPS = 104^\circ$ and angle $PQR = 87^\circ$.

Kai writes:

'Angle $QRS = 104^\circ$ because opposite angles in a cyclic quadrilateral are equal.'

(a) Explain what is wrong with Kai's reason. [1]

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

(a) opposite angles of a cyclic quadrilateral add up to $180^\circ$ (they are not equal) — must refer to adding to $180^\circ$B1
(b) $76^\circ$B1
(c) $93^\circ$B1
Answer: (a) Opposite angles of a cyclic quadrilateral add up to $180^\circ$, not equal. (b) $76^\circ$ (c) $93^\circ$
Examiner tip: In (a), 'Kai is wrong' or 'it should be $76^\circ$' is not an explanation — state the correct fact: opposite angles sum to $180^\circ$.

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