OCR GCSE Mathematics: Circle theorems and proofs (J560) – Revision Notes

These revision notes condense section 8.05 (Circles) of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026). The terms in 8.05a are for both tiers. The theorems 8.05b to 8.05h are in the Higher tier column only, so they are marked Higher tier only here. For full explanations and every proof in detail, read the study guide.

Related pages: the OCR GCSE Mathematics hub, the printable checklist, the practice set and the free diagnostics.

8.05a Terms to know (both tiers)

Term	One-line definition
Centre	Point equally far from the whole circumference
Radius	Centre to circumference
Diameter	Chord through the centre (twice the radius)
Chord	Straight line joining two points on the circumference
Circumference	The curve itself, or its length
Tangent	Line touching the circle at exactly one point
Arc	Part of the circumference
Sector	Region bounded by two radii and an arc
Segment	Region bounded by a chord and an arc

Which tier needs what

In the specification, column 1 (initial learning) names the centre, radius, chord, diameter and circumference. Column 2, for both tiers, adds tangent, arc, sector and segment. Everything from 8.05b to 8.05h is in column 3, the Higher tier only column. If you are sitting Foundation tier, learn the vocabulary above and use the rest of the page as extension work.

The seven theorems at a glance (Higher tier only)

Ref	Theorem	Reason to write
8.05b	Angle at centre = 2 × angle at circumference, same arc	“Angle at centre is twice the angle at the circumference”
8.05c	Angle in a semicircle = 90°	“Angle in a semicircle”
8.05d	Angles in the same segment are equal	“Angles in the same segment”
8.05e	Radius bisects chord if and only if perpendicular to it	“Perpendicular from the centre bisects the chord” (or the converse)
8.05f	Radius meets tangent at 90°	“Radius is perpendicular to the tangent”
8.05g	Tangent-chord angle = angle in alternate segment	“Alternate segment theorem”
8.05h	Opposite angles of a cyclic quadrilateral add to 180°	“Opposite angles of a cyclic quadrilateral”

Give the reason in words beside every angle you find.

Spotting the theorem

Diameter drawn, or a line through the centre?   -> 8.05c or 8.05e
Two radii form a triangle?                      -> isosceles, base angles equal
Angles at the centre and the edge on one arc?   -> 8.05b
Two edge angles on one chord, same side?        -> 8.05d
Four points on the circle forming a shape?      -> 8.05h
A tangent line in the diagram?                  -> 8.05f first, then 8.05g

Method in steps

1. Mark every given angle and every right angle you can justify.
2. Mark equal radii and shade the isosceles triangles they make.
3. Choose the theorem that links a known angle to the unknown angle.
4. Write the angle with its reason. Repeat until you reach the target.
5. Check: do the angles on a straight line, in a triangle or in a quadrilateral still add up?

Reminders with worked figures

Reflex and non-reflex (8.05b). The angle at the circumference on the major arc is 58°. Then the angle at the centre on the same minor arc is 2 × 58° = 116°. A point on the minor arc sees the chord at 180° - 58° = 122°, because the two points on opposite arcs and the chord ends form a cyclic quadrilateral (8.05h). The reflex angle at the centre is 360° - 116° = 244° = 2 × 122°.

Perpendicular and half-chord (8.05e). Radius 5 cm, chord 8 cm. Half-chord = 4 cm. Distance from centre: √(25 - 16) = 3 cm.

Tangent at a point (8.05f). In triangle OPT, with the tangent PT, angle OPT = 90°. If angle POT = 57°, then angle OTP = 180° - 90° - 57° = 33°.

Alternate segment (8.05g). Tangent-chord angle 38° means the angle in the alternate segment is 38°. Do not use the angle on the same side as the 38° angle.

Cyclic quadrilateral (8.05h). Angle ABC = (2x + 15)° and angle ADC = (3x - 5)°. Then 5x + 10 = 180, so x = 34, and the angles are 83° and 97°.

Worked reasoning chain (Higher tier only)

A, B, C and D lie on a circle in that order. Angle ABC = 116° and angle ABD = 47°. The tangent at A is drawn, with T on the side of AD away from B. Find angle TAD, angle ADC, angle DAC and angle ACD.

Angle TAD = angle ABD = 47°          (alternate segment theorem)
Angle ADC = 180° - 116° = 64°         (opposite angles of a cyclic quadrilateral)
Angle DBC = 116° - 47° = 69°          (angles ABD and DBC make angle ABC)
Angle DAC = angle DBC = 69°           (angles in the same segment, chord DC)
Angle ACD = 180° - 64° - 69° = 47°    (angles in triangle ACD)

Check: angle ACD should equal angle ABD, because both stand on chord AD in the same segment. Both are 47°.

Proof-writing checklist (Higher tier only)

• State what you are proving, using the letters on the diagram.
• Name every extra line you draw.
• Put a reason beside every statement.
• Use only results you have already proved or know from earlier work, such as isosceles triangles, angles on a straight line and the angle sum of a triangle.
• Finish by writing the result you set out to prove.

Proof outlines to learn (Higher tier only)

Learn the skeleton of each proof, then practise writing it out with reasons.

Result	Skeleton
8.05b	Draw the line from the circumference point through the centre. Two isosceles triangles. Exterior angles. Add (or subtract)
8.05c	Diameter gives 180° at the centre. Halve it using 8.05b
8.05d	Both angles are half the same centre angle
8.05e	Join OA and OB. SSS for one direction. RHS for the other direction
8.05f	Suppose it is not perpendicular. The foot of the perpendicular is nearer than the radius, so it is inside the circle. Contradiction
8.05g	Draw the diameter from the point of contact. Use 90° twice, then 8.05d
8.05h	Two centre angles on the two arcs total 360°. Halve

Must-know distinctions

• Centre or circumference. Angle at the centre is double. Angle at the circumference is half.
• Tangent or chord. A tangent touches once. A chord joins two points on the circle and passes inside.
• Same segment or alternate segment. Same segment: both angles on one side of the chord (8.05d). Alternate segment: the angle on the opposite side of the chord from the tangent-chord angle (8.05g).
• One-way or both ways. 8.05e is “if and only if”, so a proof needs both directions.
• Supplementary or equal. Opposite angles in a cyclic quadrilateral add to 180° and are not equal unless both are 90°.

Quick self-test

1. What is the region between a chord and its arc called?
2. The angle at the centre standing on an arc is 112°. Find the angle at the circumference on the major arc. (Higher tier only)
3. AB is a diameter and C is on the circle. Angle BAC = 41°. Find angle ABC. (Higher tier only)
4. A chord of length 8 cm is in a circle of radius 5 cm. How far is the chord from the centre? (Higher tier only)
5. In triangle OPT, PT is a tangent, angle POT = 57°. Find angle OTP. (Higher tier only)
6. In a cyclic quadrilateral ABCD, angle DAB = 107°. Find angle BCD. (Higher tier only)
7. The angle between a tangent and a chord is 38°. What other angle is 38°? (Higher tier only)
8. A, B, C and D are on a circle, with C and D both on the major arc cut off by chord AB. Angle ACB = 29°. Find angle ADB. (Higher tier only)
9. Is the word “segment” needed by Foundation candidates, Higher candidates, or both?
10. Why is the statement “if and only if” in 8.05e important for a proof? (Higher tier only)

Answers

1. A segment.
2. 56°.
3. 49° (angle in a semicircle is 90°, then the triangle sums to 180°).
4. 3 cm.
5. 33°.
6. 73°.
7. The angle in the alternate segment, on the far side of the chord.
8. 29° (angles in the same segment).
9. Both. It is in 8.05a, which is for both tiers.
10. You must prove two statements: that a radius which bisects a chord is perpendicular to it, and that a radius perpendicular to a chord bisects it.

Where marks are usually lost

• Giving a numerical answer with no reason, when the question says “give a reason”.
• Writing a vague reason such as “circle theorem” instead of naming the theorem.
• Treating a tangent as a chord, or missing the 90° at the point of contact.
• Using the alternate segment theorem with the angle in the same segment instead of the opposite one.
• Missing isosceles triangles made from two radii.
• In a proof, starting from the answer or assuming the result.
• Proving only one direction of 8.05e.
• Forgetting the reflex angle at the centre when the centre lies outside the angle.

Official syllabus

OCR, GCSE (9-1) Mathematics J560, Version 2.1 (August 2026), for first assessment in 2017: section 8.05 Circles.