IGCSE Maths circle theorems: every theorem, with the reasons examiners accept

Which theorems each tier needs

From Cambridge 0580 C4.7, E4.7, E4.8 and Edexcel 4MA1 4.6A to 4.6C.

The key ideas

Most circle questions are solved by spotting a hidden shape. Two radii make an isosceles triangle, so its base angles are equal. A tangent and a radius make a right angle. Four points on the circle make a cyclic quadrilateral. A diameter means a right angle somewhere on the circumference.

The angle at the centre theorem says that the angle subtended by an arc at the centre is twice the angle it subtends at any point on the remaining circumference. The semicircle theorem is a special case: a diameter makes 180° at the centre, so 90° at the circumference.

Angles in the same segment are equal: two angles standing on the same chord, on the same side of it, are equal. Opposite angles of a cyclic quadrilateral add to 180°. The alternate segment theorem says the angle between a tangent and a chord equals the angle in the alternate segment, the angle on the far side of the chord.

Edexcel Higher also tests intersecting chords. If chords AB and CD cross at P inside the circle, AP × PB = CP × PD. If they meet outside at P, PA × PB = PC × PD, and for a tangent PT from P, PT^2 = PA × PB.

Worked example 1: angle at the centre and an isosceles triangle

1. A, B and C lie on a circle with centre O. Angle ACB = 38°. Find angle AOB and angle OAB.
2. Angle AOB = 2 × 38° = 76°. Reason: the angle at the centre is twice the angle at the circumference.
3. OA = OB because both are radii, so triangle AOB is isosceles.
4. Angle OAB = (180° - 76°)/2 = 52°. Reason: base angles of an isosceles triangle are equal and angles in a triangle sum to 180°.

Worked example 2: tangents from an external point

1. TA and TB are tangents from T to a circle with centre O. Angle ATB = 50°. Find angle AOB, then angle ACB where C is on the major arc.
2. Angle OAT = angle OBT = 90°. Reason: a tangent is perpendicular to the radius at the point of contact.
3. Angle AOB = 360° - 90° - 90° - 50° = 130°. Reason: angles in a quadrilateral sum to 360°.
4. Angle ACB = 130°/2 = 65°. Reason: the angle at the centre is twice the angle at the circumference.

Worked example 3: cyclic quadrilateral and alternate segment

1. ABCD is a cyclic quadrilateral with angle A = (2x + 30)° and angle C = (3x - 10)°. Opposite angles sum to 180°, so 5x + 20 = 180, x = 32, angle A = 94° and angle C = 86°. Check: 94 + 86 = 180.
2. Separately, a tangent touches a circle at A. The angle between the tangent and chord AB is 64°. C is a point on the circle in the alternate segment.
3. Angle ACB = 64°. Reason: alternate segment theorem.
4. Edexcel Higher extra: chords AB and CD meet at P inside a circle with AP = 6, PB = 4 and CP = 3. Then 6 × 4 = 3 × PD, so PD = 8.

Common mistakes that cost marks

• Giving a vague reason such as 'circle theorem' or 'because of the tangent'. Use the full standard wording.
• Halving instead of doubling, or the reverse, in the angle at the centre theorem. The centre angle is the larger one.
• Using the cyclic quadrilateral rule on a quadrilateral whose four vertices are not all on the circle, for example one with a vertex at the centre.
• Applying the semicircle theorem when the chord is not a diameter (it does not pass through the centre).
• Choosing the wrong angle in the alternate segment theorem; it is the angle on the opposite side of the chord from the tangent angle.
• Assuming a diagram is accurate. Diagrams are not drawn to scale, so never measure.

Exam technique and how a tutor helps

On both boards, an angle with no reason often scores only part of the marks. Write each step as an angle, its value and the reason in standard wording, for example 'angle OBT = 90°, tangent perpendicular to radius'. Mark every angle you find on the diagram as you go, because the next step usually depends on it.

When you are stuck, look for radii (isosceles triangles), diameters (right angles) and tangents (right angles with radii). Those three spot most of the hidden structure.

A one-to-one tutor helps most with seeing the structure, which is a visual skill. On the shared whiteboard the tutor can highlight the isosceles triangle or the cyclic quadrilateral in colour, then have the student find the next one themselves. Lessons also drill the reason wording until it is automatic, since that is where many marks are lost.

Self-check: can you do these?

• An angle at the circumference is 47°. What is the angle at the centre on the same arc? (Answer: 94°)
• AB is a diameter and C is on the circle. Angle CAB = 35°. Find angle CBA. (Answer: 55°)
• Two opposite angles of a cyclic quadrilateral are x and 2x. Find x. (Answer: 60°)
• Tangents from T meet the circle at A and B. TA = 9 cm. What is TB? (Answer: 9 cm)
• Write the standard reason for the angle between a tangent and a radius.