Skip to main content

O-Level Mathematics · Secondary 3 and 4

Angle Properties of Circles

Circle geometry is about the angles and lengths that appear when chords, radii and tangents are drawn in a circle. In O-Level Mathematics you use the symmetry properties of chords and tangents and the angle properties of circles to find unknown angles, giving a geometric reason for each step. These questions often combine circle facts with isosceles triangles, parallel lines and similar triangles, so knowing each property and its exact wording is the key to working through them.

What the syllabus expects

  • Use the symmetry properties of chords: the perpendicular from the centre to a chord bisects the chord, and equal chords are equidistant from the centre
  • Use the tangent properties: a tangent is perpendicular to the radius at the point of contact, and tangents from an external point are equal in length
  • Use the fact that the angle in a semicircle is a right angle
  • Use the fact that the angle at the centre is twice the angle at the circumference
  • Use the facts that angles in the same segment are equal and that opposite angles of a cyclic quadrilateral add up to 180°
  • Solve angle problems with clear reasons, combining circle properties with triangle and parallel-line facts

Symmetry properties

Every radius of a circle is the same length, so any triangle formed by two radii and a chord is isosceles. This simple fact is used in nearly every question. The perpendicular from the centre to a chord bisects the chord, and the line from the centre to the midpoint of a chord is perpendicular to it. This creates right-angled triangles, so you can use Pythagoras' theorem. For a circle of radius 13 cm and a chord 24 cm long, half the chord is 12 cm and the distance from the centre to the chord is √(13² − 12²) = √25 = 5 cm. Equal chords are the same distance from the centre, and chords that are the same distance from the centre are equal in length.

Tangent properties

A tangent touches the circle at exactly one point, and it is perpendicular to the radius drawn to that point, so the angle between them is 90°. From a point P outside the circle, two tangents can be drawn, touching the circle at A and B. These tangents are equal in length, PA = PB, and the line joining P to the centre O bisects angle APB and angle AOB. This creates two congruent right-angled triangles, OAP and OBP. In the kite OAPB, the angles at A and B are both 90°, so angle AOB + angle APB = 180°. For example, if angle APB = 40°, then angle AOB = 140°. When you use these facts, write the reasons as 'tangent perpendicular to radius' and 'tangents from an external point are equal'.

Angles at the centre and in a semicircle

The angle subtended by an arc at the centre is twice the angle it subtends at any point on the remaining part of the circumference. If the minor arc AC subtends 110° at the centre O, then at a point B on the major arc, angle ABC = 55°. The rule also works with a reflex angle at the centre. The major arc AC subtends the reflex angle 360° − 110° = 250° at O, so at a point D on the minor arc, angle ADC = 125°. Notice that 55° + 125° = 180°, which is where the cyclic quadrilateral property comes from. The angle in a semicircle is a special case. A diameter subtends 180° at the centre, so it subtends 90° at any point on the circumference. Look for a diameter in every diagram, because it gives you a right angle.

Same segment and cyclic quadrilaterals

Angles subtended by the same arc at points on the circumference, on the same side of the chord, are equal. These are called angles in the same segment. In a diagram with chords crossing inside a circle, look for two angles that stand on the same arc, often forming a bow-tie shape. A cyclic quadrilateral has all four vertices on the circle, and its opposite angles add up to 180°. So if angle ABC = 70° in cyclic quadrilateral ABCD, then angle ADC = 110°. A quadrilateral with one vertex at the centre of the circle is not cyclic, so this property does not apply to it. When writing reasons, use the standard phrases: 'angle at centre = 2 × angle at circumference', 'angles in the same segment', 'angles in opposite segments' and 'angle in a semicircle'.

Worked example

A, B, C and D lie on a circle with centre O. B lies on the major arc AC and D lies on the minor arc AC, so ABCD is a cyclic quadrilateral. Angle AOC = 140°, where this is the angle facing the minor arc. Find (a) angle ABC, (b) angle ADC, (c) angle OAC, giving reasons.

  1. Angle ABC stands on the minor arc AC, which subtends 140° at the centre, so angle ABC = 140° ÷ 2 = 70° (angle at centre = 2 × angle at circumference).
  2. ABCD is a cyclic quadrilateral, so angle ADC = 180° − 70° = 110° (angles in opposite segments).
  3. Check with the reflex angle: reflex angle AOC = 360° − 140° = 220°, and angle ADC = 220° ÷ 2 = 110°, which agrees.
  4. Triangle AOC is isosceles because OA = OC (radii).
  5. So angle OAC = (180° − 140°) ÷ 2 = 20° (base angles of an isosceles triangle).

Answer: (a) 70° (b) 110° (c) 20°

Common mistakes

  • Halving the wrong angle at the centre. The angle at the centre and the angle at the circumference must stand on the same arc. If the point on the circumference is on the minor arc, use the reflex angle at the centre.
  • Assuming a line is a diameter. The angle in a semicircle is 90° only if the chord passes through the centre. Do not assume this from the picture unless the question says so or O clearly lies on the line.
  • Using the cyclic quadrilateral rule on a shape that is not cyclic. A quadrilateral with a vertex at the centre, such as OABC, does not have all four vertices on the circle, so its opposite angles need not add up to 180°.
  • Forgetting the isosceles triangles. Two radii always make an isosceles triangle, and many missing angles come from equal base angles rather than from a circle property.
  • Writing vague reasons. 'Circle property' does not explain anything. Name the exact property, such as 'angles in the same segment' or 'tangent perpendicular to radius'.
  • Using the right angle between tangent and radius at the wrong point. The 90° is only at the point of contact, between the tangent and the radius drawn to that point.

Exam tips

  • Mark every radius on the diagram as you notice it. Equal radii reveal isosceles triangles, which are often the missing link.
  • Write each angle with its reason on the same line, such as 'angle ADC = 110° (angles in opposite segments)'. The reasons are part of the answer in these questions.
  • If you are stuck, look for a diameter, a tangent, or two angles standing on the same arc. Spotting these three features is the way into most circle questions.

Practice questions

1. AB is a diameter of a circle and C is a point on the circumference. Angle CAB = 35°. Find angle ABC.

Show answer

55°. Angle ACB = 90° (angle in a semicircle), so angle ABC = 180° − 90° − 35° = 55° (angle sum of a triangle).

2. PA and PB are tangents to a circle with centre O, touching it at A and B. Angle APB = 50°. Find angle AOB and angle PAB.

Show answer

Angle AOB = 130° and angle PAB = 65°. Angles OAP and OBP are 90° (tangent perpendicular to radius), so angle AOB = 360° − 90° − 90° − 50° = 130°. PA = PB (tangents from an external point), so triangle PAB is isosceles and angle PAB = (180° − 50°) ÷ 2 = 65°.

3. A, B, C and D lie on a circle in that order, and the chords AC and BD meet at X. Angle BAC = 32° and angle ABD = 45°. Find angle BDC, angle ACD and angle AXB.

Show answer

Angle BDC = angle BAC = 32°, because both stand on arc BC (angles in the same segment). Angle ACD = angle ABD = 45°, because both stand on arc AD. In triangle ABX, angle AXB = 180° − 32° − 45° = 103°.

Related topics

Want a tutor for E-Math?

Shortlist on WhatsApp within 16 hours. You choose the tutor. About E-Math tuition

Get matched with a tutor

Or ask on WhatsApp