Circles

Q: In a circle with centre O, ∠ AOB = 96° and C lies on the major arc AB. Find ∠ ACB.

∠ ACB = (1)/(2) × 96° = 48°.

Q: AB is a diameter of a circle and C is a point on it. If ∠ CAB = 37°, find ∠ ABC, giving a reason.

∠ ACB = 90° (angle in a semicircle), so ∠ ABC = 180° - 90° - 37° = 53°.

Q: The side DC of a cyclic quadrilateral ABCD is produced to E, and ∠ BCE = 80°. Find ∠ BAD.

∠ BAD = 80° (exterior angle of a cyclic quadrilateral equals the interior opposite angle).

Q: A, B, C, D lie on a circle and C, D are on the same side of the chord AB. If ∠ ADB = 42°, find ∠ ACB.

∠ ACB = 42° (angles in the same segment).

ICSE Class 10 · Mathematics · Chapter 17

What this chapter covers

Arcs, segments and the angles they subtend
The angle at the centre is double the angle at the circumference
Angles in the same segment are equal
The angle in a semicircle is a right angle
Opposite angles of a cyclic quadrilateral are supplementary
The exterior angle of a cyclic quadrilateral
Using the angle properties together
Common mistakes and how to avoid them

Key points

The angle that an arc subtends at the centre is double the angle it subtends at any point on the remaining part of the circle: ∠ AOB = 2∠ ACB.
For a point on the minor arc, use the reflex angle at the centre.
Angles in the same segment of a circle are equal (without proof in the syllabus).
The angle in a semicircle is 90^°, because the diameter makes 180^° at the centre.
The opposite angles of a cyclic quadrilateral add up to 180^°.
The exterior angle of a cyclic quadrilateral equals the interior opposite angle (without proof in the syllabus).
Two radii always make an isosceles triangle, with equal base angles.
Write the reason with every step, using the standard phrase for the property.

Questions answered

Define a cyclic quadrilateral.

A cyclic quadrilateral is a quadrilateral whose four vertices all lie on one circle.

In a circle with centre O, ∠ AOB = 96^° and C lies on the major arc AB. Find ∠ ACB.

∠ ACB = (1)/(2) × 96^° = 48^°.

C lies on the minor arc AB of a circle with centre O, and ∠ ACB = 124^°. Find the angle ∠ AOB that is not reflex.

C is on the minor arc, so ∠ ACB is half the reflex angle: reflex ∠ AOB = 248^°, so ∠ AOB = 360^° - 248^° = 112^°.

AB is a diameter of a circle and C is a point on it. If ∠ CAB = 37^°, find ∠ ABC, giving a reason.

∠ ACB = 90^° (angle in a semicircle), so ∠ ABC = 180^° - 90^° - 37^° = 53^°.

In a cyclic quadrilateral ABCD, ∠ A = (2x + 4)^° and ∠ C = (4x - 22)^°. Find x, ∠ A and ∠ C.

2x + 4 + 4x - 22 = 180, so 6x = 198 and x = 33. Then ∠ A = 70^° and ∠ C = 110^°.

The side DC of a cyclic quadrilateral ABCD is produced to E, and ∠ BCE = 80^°. Find ∠ BAD.

∠ BAD = 80^° (exterior angle of a cyclic quadrilateral equals the interior opposite angle).

Why must a cyclic parallelogram be a rectangle?

Its opposite angles are equal because it is a parallelogram and supplementary because it is cyclic, so each is 90^°; a parallelogram with right angles is a rectangle.

A, B, C, D lie on a circle and C, D are on the same side of the chord AB. If ∠ ADB = 42^°, find ∠ ACB.

∠ ACB = 42^° (angles in the same segment).

Past board exam questions

ICSE 2026 sample paper · 4 marks

AC is the diameter of the circle with centre O. Points C, B, A, E and D lie on the circle in that order, so B is on one side of AC and E and D are on the other. OB, CB, CE, BE, CD and DE are joined. CD is parallel to BE. ∠AOB = 80° and ∠ACE = 20°. Calculate: (a) ∠ BEC (b) ∠ BCD (c) ∠CED

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(a) ∠BOC = 180° − ∠AOB = 180° − 80° = 100° (AOC is a straight line). ∠BEC = (1)/(2)∠BOC = 50° (angle at the centre is twice the angle at the remaining part of the circle). (b) ∠BCA = (1)/(2)∠AOB = 40°. ∠ECD = ∠BEC = 50° (alternate angles, CD ∥ BE). ∠BCD = ∠BCA + ∠ACE + ∠ECD = 40° + 20° + 50° = 110°. (c) BCDE is a cyclic quadrilateral, so ∠BED = 180° − ∠BCD = 70°. ∠CED = ∠BED − ∠BEC = 70° − 50° = 20°.

ICSE 2026 board paper · 5 marks

Use ruler and compass for the following construction: (a) construct an equilateral triangle ABC of side 5 cm. (b) construct the circumcircle of ΔABC. (c) construct the locus of points which are equidistant from AB and BC. Mark the point where the circumcircle and locus meet, as D. (d) give the geometrical name of quadrilateral ABCD.

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(a) Draw BC = 5 cm. With B and C as centres and radius 5 cm, draw arcs meeting at A. Join AB and AC. (b) Draw the perpendicular bisectors of AB and BC; they meet at O, the circumcentre. With centre O and radius OA (= (5)/(√(3)) ≈ 2.9 cm) draw the circumcircle through A, B and C. (c) The locus of points equidistant from AB and BC is the bisector of ∠ ABC. Draw it; it passes through O and meets the circumcircle again at D (the point diametrically opposite B). (d) BD is a diameter and the perpendicular bisector of AC, so AB = BC and AD = DC (with ∠ BAD = ∠ BCD = 90^°, ∠ ADC = 120^°). ABCD is a kite (a cyclic kite).

ICSE 2025 board paper · 4 marks

Use ruler and compass for the following construction. Construct a ΔABC, where AB = 6 cm, AC = 4.5 cm and ∠BAC = 120°. Construct a circle circumscribing the ΔABC. Measure and write down the length of the radius of the circle.

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Steps: (1) Draw AB = 6 cm. (2) At A construct ∠ BAC = 120^° and cut AC = 4.5 cm; join BC. (3) Draw the perpendicular bisectors of AB and AC; they meet at O (outside the triangle, since the triangle is obtuse). (4) With centre O and radius OA, draw the circle; it passes through A, B and C. Radius of the circumcircle ≈ 5.3 cm. (Check: BC² = 6² + 4.5² - 2(6)(4.5)cos 120^° = 83.25, BC ≈ 9.12 cm; R = (BC)/(2 sin 120^°) ≈ 5.27 cm.)

ICSE 2025 compartment paper · 4 marks

Construct a regular hexagon of side 4.5 cm. Hence, construct a circle circumscribing the regular hexagon. Use ruler and compass for the construction. Measure and write down the radius of the circle.

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Steps of construction: 1. Draw AB = 4.5 cm. 2. At A and B construct angles of 120° (each interior angle of a regular hexagon is 120°) and cut off AF = BC = 4.5 cm. 3. Similarly, at C and F construct 120° angles and cut off CD = FE = 4.5 cm; join DE. ABCDEF is the regular hexagon. 4. Draw the perpendicular bisectors of AB and BC; they meet at O. 5. With centre O and radius OA, draw the circle; it passes through all six vertices. Radius of the circumscribing circle = 4.5 cm (in a regular hexagon each side subtends 60° at the centre, so the circumradius equals the side).

ICSE 2024 sample paper · 4 marks

O is the centre of circle circumscribing the ∆ABC. OA and OC are joined. CD is perpendicular to chord AB, with D on AB. ∠OAC=32°. Find each of the unknown angles x, y and z, where x° = ∠AOC, y° = ∠ABC and z° = ∠DCB.

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OA = OC (radii), so ∠OCA = ∠OAC = 32°. In ∆OAC, x = ∠AOC = 180° − 32° − 32° = 116°. y = ∠ABC = (1)/(2)∠AOC = (1)/(2) × 116° = 58° (angle at the centre is twice the angle at the remaining part of the circle). In ∆BDC, ∠BDC = 90°, so z = ∠DCB = 180° − 90° − 58° = 32°. x = 116°, y = 58°, z = 32°.

ICSE 2023 board paper · 4 marks

Using ruler and compass construct a triangle ABC in which AB = 6 cm, ∠BAC = 120° and AC = 5 cm. Construct a circle passing through A, B and C. Measure and write down the radius of the circle.

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Steps of construction: 1. Draw AB = 6 cm. 2. At A, construct ∠BAC = 120° with ruler and compass and cut AC = 5 cm on its arm. Join BC. 3. Draw the perpendicular bisectors of AB and AC; they meet at O, the circumcentre. 4. With centre O and radius OA, draw the circle; it passes through A, B and C. Radius = OA ≈ 5.5 cm. (Check: BC² = 36 + 25 + 30 = 91, BC ≈ 9.54 cm; R = (BC)/(2sin 120°) ≈ 5.51 cm.)

ICSE 2023 sample paper · 3 marks

Using ruler and compasses, construct a regular hexagon of side 4.5 cm. Hence construct a circle circumscribing the hexagon. Measure and write down the length of the circum-radius.

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Steps of construction: 1. Draw AB = 4.5 cm. 2. At A and B construct angles of 120° and cut off AF = BC = 4.5 cm. 3. With F and C as centres and radius 4.5 cm, draw arcs; at F and C construct 120° angles to locate E and D with FE = CD = ED = 4.5 cm. ABCDEF is the regular hexagon. 4. Draw the perpendicular bisectors of AB and BC; they meet at O (equivalently, O is where the diagonals AD, BE, CF meet). 5. With centre O and radius OA, draw the circle; it passes through all six vertices. Circum-radius = OA = 4.5 cm (in a regular hexagon the circum-radius equals the side).

ICSE 2022 sample paper · 1 mark

AB is a diameter of the circle with centre 'O'. C is a point on the circle; the chord AC and the radius OC are drawn, and x is the angle ACO between them. If ∠ COB = 55^° then the value of x is:

27.5^°
55^°
110^°
125^°

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27.5^°