Circle Theorems

This lesson is Higher tier throughout. The circle theorems are a set of rules about the angles formed by radii, chords, tangents and points on a circle, and they are a reliable source of marks on the Higher papers of OCR GCSE Mathematics (J560). Every question is really the same task: spot which theorem applies, use it to find an angle, and — crucially — give the theorem by name as your reason. The reasoning is worth as much as the arithmetic here. Recalling a theorem is AO1; selecting and naming the right one in a diagram is AO2; and chaining several together to reach an answer is AO3. Learn the theorems with their standard wording, because that wording is exactly what earns the reason marks.

Key Vocabulary

Term	Meaning
Chord	A straight line joining two points on the circumference
Arc	A part of the circumference
Subtend	To "stand on" — an arc or chord subtends the angle opening onto it
Cyclic quadrilateral	A quadrilateral with all four vertices on the circle
Tangent	A line touching the circle at exactly one point
Segment	The region between a chord and an arc
Semicircle	Half a circle, cut off by a diameter

The Seven Circle Theorems

There are seven theorems to know. (1) The angle at the centre is twice the angle at the circumference when both stand on the same arc. (2) The angle in a semicircle is $90^{\circ}$ . (3) Angles in the same segment (standing on the same arc) are equal. (4) Opposite angles of a cyclic quadrilateral sum to $180^{\circ}$ . (5) A tangent is perpendicular to the radius at the point of contact. (6) Two tangents from the same external point are equal in length. (7) The alternate segment theorem: the angle between a tangent and a chord equals the angle in the alternate segment. The most quoted of these is the first, so its diagram is shown here.

Angle at the Centre and Angle in a Semicircle

The first theorem is the one you will use most, so it deserves careful attention. It says the angle $\angle AOB$ formed at the centre of the circle is exactly twice the angle $\angle ACB$ formed at the circumference, provided both angles are subtended by — that is, stand on — the same arc $AB$ . The word "subtended" simply means the arc opens onto the angle; both the centre angle and the circumference angle must "look at" the same arc for the relationship to hold. A neat way to picture it is that the point $C$ on the circumference sees the arc from a shallower viewpoint than the centre does, and that shallower view is always precisely half. The second theorem is just a special case of the first, and recognising this saves you memorising it separately. If the chord $AB$ happens to be a diameter, then the angle it makes at the centre is the straight angle $180^{\circ}$ ; halving that gives $90^{\circ}$ , so the angle subtended at the circumference by a diameter is always a right angle. This is the famous angle in a semicircle, and it means that any triangle drawn with one side as the diameter and its third vertex on the circle must be right-angled at that vertex.

Worked Example 1

$O$ is the centre of a circle. $A$ and $B$ lie on the circle and $\angle AOB = 124^{\circ}$ . Point $C$ is on the major arc. Work out $\angle ACB$ , giving a reason.

The angle at the centre is twice the angle at the circumference on the same arc, so $\angle ACB = \dfrac{124^{\circ}}{2} = 62^{\circ}$ (angle at centre is twice angle at circumference).

Answer: $62^{\circ}$ .

Worked Example 2

$PR$ is a diameter of a circle with centre $O$ , and $Q$ is another point on the circle. $\angle QPR = 28^{\circ}$ . Work out $\angle PQR$ and $\angle QRP$ , with reasons.

Since $PR$ is a diameter, $\angle PQR = 90^{\circ}$ (angle in a semicircle). Then in $\triangle PQR$ the angles sum to $180^{\circ}$ , so $\angle QRP = 180^{\circ} - 90^{\circ} - 28^{\circ} = 62^{\circ}$ .

Answer: $\angle PQR = 90^{\circ}$ , $\angle QRP = 62^{\circ}$ .

Common error: halving the wrong angle, or applying the "twice" rule when the two angles are not on the same arc.

Angles in the Same Segment

The third theorem says that two angles standing on the same arc, from different points on the same side of a chord, are equal — they are called "angles in the same segment". The reason follows directly from the first theorem: both circumference angles are half of the same centre angle, so they must equal each other. In practice this theorem is extremely useful for transferring a known angle to a new position in the diagram: if you know the angle one point makes on a chord, you instantly know the angle any other point on the same arc makes on it. A common diagram has a chord $AB$ at the bottom and two points $C$ and $D$ further round the circle, with the angles $\angle ACB$ and $\angle ADB$ both opening onto $AB$ ; these two angles are equal. The figure shows exactly this configuration.

Worked Example 3

Points $A$ , $B$ , $C$ , $D$ lie on a circle. $\angle ACB = 35^{\circ}$ , where $C$ and $D$ are on the same arc relative to chord $AB$ . Work out $\angle ADB$ , with a reason.

Angles in the same segment are equal, so $\angle ADB = \angle ACB = 35^{\circ}$ (angles in the same segment are equal).

Answer: $35^{\circ}$ .

Cyclic Quadrilaterals

The fourth theorem concerns a cyclic quadrilateral, which is simply a four-sided shape whose every vertex lies on the circle — the quadrilateral is "inscribed" in the circle. The theorem is short and powerful: its opposite angles sum to $180^{\circ}$ . This means the two angles at opposite corners are supplementary, so knowing one of a pair instantly gives the other as $180^{\circ}$ minus it. The crucial detail, and the one most often slipped on, is that the rule links opposite corners, not adjacent ones; in a quadrilateral labelled $ABCD$ in order around the circle, it pairs $A$ with $C$ and $B$ with $D$ . It is worth remembering that a quadrilateral only has this property if all four vertices truly lie on the circle — if even one does not, the rule does not apply. The figure shows a cyclic quadrilateral inscribed in a circle.

Worked Example 4

$ABCD$ is a cyclic quadrilateral with $\angle A = 95^{\circ}$ and $\angle B = 70^{\circ}$ . Work out $\angle C$ and $\angle D$ , with reasons.

Opposite angles of a cyclic quadrilateral sum to $180^{\circ}$ . So $\angle C = 180^{\circ} - 95^{\circ} = 85^{\circ}$ (opposite to $\angle A$ ) and $\angle D = 180^{\circ} - 70^{\circ} = 110^{\circ}$ (opposite to $\angle B$ ).

Answer: $\angle C = 85^{\circ}$ , $\angle D = 110^{\circ}$ .

Common error: pairing adjacent angles instead of opposite ones — the theorem links $A$ with $C$ and $B$ with $D$ .

Circle Theorems (Higher)