Circle theorems II: IGCSE Mathematics questions with answers
Free Circle theorems II questions for IGCSE Mathematics (0580), Extended tier (Supplement content), part of Geometry. Each question has the answer and the marking points an examiner would look for.
A, B, C and D lie on a circle, centre O. Angle AOC = 140°.
Fig. 1.1
(a)
Find angle ABC. Give a reason for your answer.
[2]
(b)
Find angle ADC. Give a reason for your answer.
[2]
(c)
The tangent to the circle at C is drawn. Find the angle between this tangent and the chord CA, on the same side of CA as D. Give a reason.
[2]
(d)
Find angle OAC.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
70° (angle at centre is twice the angle at the circumference)
70 B1 angle at centre = 2 × angle at circumference B1
2
(b)
110° (opposite angles of a cyclic quadrilateral add up to 180°)
110 B1 opposite angles of a cyclic quadrilateral sum to 180° B1
2
(c)
70° (alternate segment theorem)
70 B1 alternate segment theorem B1
2
(d)
20°
20 (isosceles triangle OAC) B1
1
Question 2Structured · 11 marksA* challenge
The n vertices of a regular polygon lie on a circle with centre O. Three of the vertices, P, Q and R, are joined to form triangle PQR.
(a)
Explain why each angle of triangle PQR is a multiple of 180n degrees.
[3]
(b)
Angle PQR = 99°. Find the smallest possible value of n.
[3]
(c)
Angle PQR = 99° and triangle PQR is isosceles. Find the smallest possible value of n, and the other two angles of the triangle.
[3]
(d)
n = 20 and angle PQR = 99°. Find the number of non-congruent triangles PQR that are possible.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
each side of the polygon subtends 360n° at O; an angle of the triangle is half the angle at O subtended by k sides
each side of the polygon subtends 360n° at the centre B1 angle at the circumference is half the angle at the centre (same arc) M1 arc opposite the angle contains k sides, so angle = k × 180nA1
3
(b)
n = 20
99 = 180knM1 n = 20k11, so k is a multiple of 11 M1 n = 20 (k = 11, the other arcs total 9 sides) A1
3
(c)
n = 40; 40.5° and 40.5°
the other two angles must be equal: 40.5° each B1 40.5 = 180jn needs n a multiple of 40, and 99° needs n a multiple of 20 M1 n = 40 A1
3
(d)
4
arc PR (not containing Q) has 11 sides, so arcs PQ and QR together contain 9 sides M1 splits 1 + 8, 2 + 7, 3 + 6, 4 + 5: 4 triangles A1
2
Question 3Structured · 2 marks
A, B, C and D are points on the circle.
Fig. 3.1
Find the value of y. Give a reason for your answer.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
y = 60
60 B1 Angles in the same segment are equal B1
2
Question 4Structured · 8 marksA* challenge
A, B, C and D lie on a circle in that order, so ABCD is a cyclic quadrilateral. TA is the tangent to the circle at A. Angle ABC = (3x + 10)°, angle ADC = (110 − x)° and angle TAB = 50°, where T is on the opposite side of AB from C.
(a)
Find the value of x. Give a reason for each step.
[3]
(b)
Find angle ACB. Give a reason.
[2]
(c)
Find angle BAC. Give reasons.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
x = 30
opposite angles of a cyclic quadrilateral add up to 180° B1 3x + 10 + 110 − x = 180 M1 x = 30 A1
3
(b)
50°
50° B1 alternate segment theorem B1
2
(c)
30°
angle ABC = 100° B1 angles in triangle ABC add to 180° M1 30° A1
3
Question 5Structured · 7 marksA* challenge
A, B, C and D lie on a circle, centre O. Angle AOC = 136°.
Fig. 5.1
(a)
Find angle ABC. Give a reason for your answer.
[2]
(b)
Find angle ADC. Give a reason for your answer.
[2]
(c)
The tangent to the circle at C is drawn. Find the angle between this tangent and the chord CA, on the same side of CA as D. Give a reason.
[2]
(d)
Find angle OAC.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
68° (angle at centre is twice the angle at the circumference)
68 B1 angle at centre = 2 × angle at circumference B1
2
(b)
112° (opposite angles of a cyclic quadrilateral add up to 180°)
112 B1 opposite angles of a cyclic quadrilateral sum to 180° B1
2
(c)
68° (alternate segment theorem)
68 B1 alternate segment theorem B1
2
(d)
22°
22 (isosceles triangle OAC) B1
1
Question 6Structured · 2 marks
A, B, C and D are points on the circle. ABCD is a cyclic quadrilateral.
Fig. 6.1
Find the value of y. Give a reason for your answer.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
y = 121
121 B1 Opposite angles of a cyclic quadrilateral add up to 180° B1
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in