Circle Theorems

Mathematics revision notes · Cambridge (CIE) · 0580

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1. How circle questions work

Reasons carry the marks

Many circle questions ask for an angle and the theorem that justifies it, with a mark for each
On some questions the reason is a whole part on its own, so the wording has to be right
Where the answer takes two steps, two reasons are expected: give one for each step, not just the last one
The ordinary angle facts stay available throughout: angles in a triangle, angles on a line, angles at a point, isosceles triangles and angles in a quadrilateral
Mark the radii on a diagram first, since any two of them make an isosceles triangle and that is often the missing step

A circle with centre O and points A, B and C on it, all joined to O and to each other: the angle BOC at the centre is 150 degrees and the angle at B is 60, and because OA, OB and OC are all radii the triangles inside are isosceles, which is what unlocks the angle x at A — Source: Angles at centre & circumference by Save My Exams

Diagrams are not to scale, so never measure

The theorems you need

Six angle theorems are examined: the angle in a semicircle, the angle between a tangent and a radius, the angle at the centre, angles in the same segment, opposite angles of a cyclic quadrilateral, and the alternate segment theorem
Three further symmetry properties finish the list, and they are gathered in the last section of this note

2. The two right-angle theorems

The angle in a semicircle is 90°

Where a triangle has the diameter as one side and its third corner on the circumference, the angle at that corner is a right angle
To spot it, check that one side runs through the centre and that every corner lies on the circumference
Whichever corner faces the diameter is the one holding the right angle

A circle with centre O and a triangle drawn inside it: one side is a diameter running through O, the third corner sits on the circumference, and the angle at that corner is marked with a small square as a right angle — Source: Angle in a semicircle by Save My Exams

It is the special case of the centre theorem, since the straight angle of 180° at the centre halves to 90°

Worked example

Using the semicircle and the triangle

The diameter LN is drawn in a circle, and M is a third point on the circumference. Angle MLN is 28°. Find angle MNL, giving reasons.

A circle with a horizontal diameter drawn from L on the left to N on the right, and a third point M on the circumference above it. Triangle LMN is drawn, and the angle at L between the diameter and LM is marked 28 degrees. The angle at M is not labelled. The figure is marked NOT TO SCALE.

Solution:

LN is a diameter and M lies on the circumference, so angle LMN = 90°
The reason is that the angle in a semicircle is 90°
The three angles of triangle LMN add to 180°
Angle MNL = 180 − 90 − 28 = 62°
The second reason is that the angles in a triangle add to 180°

A tangent meets a radius at 90°

A tangent touches the circle at exactly one point, and the radius drawn to that point is perpendicular to it
Draw the radius in yourself whenever a tangent appears, because the right angle is usually what unlocks the question
A mark scheme has accepted "tangent and radius are perpendicular" as well as the ninety-degree form, and allowed diameter in place of radius

A circle with centre O and a straight tangent touching it at one point, with the radius drawn out to that point and a small square marking the right angle between the two — Source: Circles & chords by Save My Exams

3. The angle theorems

The angle at the centre is twice the angle at the circumference

Take a pair of points on the circumference and join each of them to the centre, and also to some third point on the circumference
Whatever angle appears at the centre is double the one appearing at that third point
The shape often looks like an arrowhead, and the two angles must be subtended by the same arc

The same theorem in a less obvious arrangement: P and Q are joined to the centre and to a third point close to P, so the shape no longer looks like an arrowhead, but the angle at the centre is still twice the angle at the circumference — Source: Angles at centre & circumference by Save My Exams

Where the centre angle is reflex, it is that reflex value which doubles the circumference angle, not the smaller one beside it

A circle where the third point sits on the short arc, so the angle at the centre facing the same arc is the reflex one, marked 2x, and it is that reflex angle rather than the smaller one beside it which is double the angle x at the circumference — Source: Angles at centre & circumference by Save My Exams

An angle of 33° at the circumference gives 66° at the centre

A circle with P and Q on the circumference and the arc between them picked out in red: both are joined to the centre, giving an angle of 2x there, and both are joined to a third point on the circumference, where the angle is x — Source: Angles at centre & circumference by Save My Exams

Angles in the same segment are equal

Two points on the circumference joined to two different third points give two equal angles
Both angles must start from the same pair of points, and both third points must lie on the same side of the chord joining them
Drawing in that chord makes the segment visible and settles whether the two angles really match

P and Q on a circle, each joined to two different points on the circumference above them, giving two equal angles marked y; the same two points are also joined to a second pair below, giving two more equal angles marked x — Source: Circles & segments by Save My Exams

Opposite angles in a cyclic quadrilateral add to 180°

A cyclic quadrilateral has all four of its corners on the circumference
Its two pairs of opposite angles each total 180°
An angle of 104° therefore sits opposite one of 76°
Check all four corners are on the circle before using it, since a quadrilateral with only three on the circumference does not qualify

A quadrilateral ABCD drawn inside a circle with centre O, every one of its four corners sitting on the circumference, which is what makes it cyclic — Source: Cyclic quadrilaterals by Save My Exams

Worked example

The centre angle and a cyclic quadrilateral

A circle has centre O, and A, B and C are points on it. Angle AOC at the centre is 130°, and B is on the major arc. (a) Find angle ABC. (b) D is a fourth point on the circle, on the minor arc AC, so ABCD is a cyclic quadrilateral. Angle BAD is 85°. Find angle BCD.

A circle with centre O marked. Points A, B and C lie on the circumference, with B on the major arc AC. The radii OA and OC are drawn and the angle between them at O, on the side away from B, is marked 130 degrees. The chords AB and BC are drawn, and the angle at B is not labelled. The figure is marked NOT TO SCALE.

Solution:

(a) Angle AOC at the centre and angle ABC at the circumference both stand on the same arc AC, so angle AOC is twice angle ABC
Angle ABC = 130 ÷ 2 = 65°
The reason is that the angle at the centre is twice any angle at the circumference standing on the same arc
(b) BAD and BCD are opposite angles of a cyclic quadrilateral, so they add to 180°
Angle BCD = 180 − 85 = 95°
The reason is "opposite angles of a cyclic quadrilateral add to 180°"
Check that all four corners really sit on the circumference before using that second theorem: a quadrilateral with one vertex at the centre, such as OABC, is not cyclic

The alternate segment theorem

The angle between a tangent and a chord equals the angle in the alternate segment, that is the angle at the circumference on the other side of that chord
Look for a triangle with all three corners on the circle and one corner touching a tangent
Find the side that forms the tangent angle, and the matching angle is the one facing it from across the triangle
An angle of 58° between tangent and chord matches one of 58° in the alternate segment
The name is the reason, so write "alternate segment theorem" in full

A triangle with all three corners on a circle, one corner resting on a tangent: the angle between the tangent and one side of the triangle is marked in red, and so is the equal angle facing it from the opposite corner of the triangle, on the far side of that chord — Source: Alternate segment theorem by Save My Exams

Exam tip

Write each reason as the whole theorem

A circle question asks for a reason in 8 of the 21 papers, and only the complete theorem earns that mark: "opposite angles of a cyclic quadrilateral add to 180°", "the angle in a semicircle is 90°", "alternate segment theorem". Schemes have refused "alternate segment" on its own, "AST" and "alternate angles" for the alternate segment theorem, and "right-angled triangle in a semicircle". "Supplementary" or "add to" in place of "sum to" is fine.

Worked example

Using the alternate segment theorem

Triangle ABC has all three corners on a circle. The straight line PAQ is the tangent at A, with P nearer to B and Q nearer to C. Angle QAC is 59° and angle ACB is 49°. Find angle BAC, giving a reason for each step.

A circle with triangle ABC drawn inside it, its three corners on the circle, A at the top. A straight line PAQ touches the circle at A, with P to the left on the side of B and Q to the right on the side of C. The angle between AQ and the chord AC is marked 59 degrees and the angle ACB at C is marked 49 degrees; angle BAC is not marked. The figure is marked NOT TO SCALE.

Solution:

Angle QAC is the angle between the tangent AQ and the chord AC
By the alternate segment theorem it equals the angle at the circumference on the other side of AC, which is angle ABC
So angle ABC = 59°, and the reason is "alternate segment theorem"
The three angles of triangle ABC add to 180°
Angle BAC = 180 − 59 − 49 = 72°
The second reason is that the angles in a triangle add to 180°
Check: the same theorem with the chord AB gives angle PAB = angle ACB = 49°, and the angles on the straight line PAQ add up: 49 + 72 + 59 = 180

4. Combining theorems

Most questions need two or three facts chained together, so work outwards from whatever is given
Fill in every angle you can and label each one with its reason as you go

A circle with five points A to E on it joined by several chords, with angles of 12, 14 and 73 degrees marked at E and the angle theta at B to be found, which needs more than one theorem chained together — Source: Circles & segments by Save My Exams

A pair of tangents drawn from one outside point, together with the two radii, boxes in a quadrilateral whose angles total 360°, and two of those are the right angles where the tangents touch

Two tangents drawn from an outside point T touching a circle of centre O at R and S, the two tangent lengths TR and TS marked equal, and the radii OR and OS drawn in to complete the quadrilateral ORTS — Source: Circles & chords by Save My Exams

Worked example

Two tangents and a quadrilateral

Two tangents from an external point T touch a circle, centre O, at R and S. Angle RTS is 25°. Find the angle θ at the centre, giving reasons.

Solution:

OR and OS are radii drawn to the points where the tangents touch
A tangent meets a radius at 90°, so angle ORT = angle OST = 90°
ORTS is a quadrilateral, and the angles of a quadrilateral add to 360°

Two tangents from a point T touching a circle of centre O at R and S, with the right angles at R and S marked and the 25 degree angle at T shown, so the four angles of ORTS add to 360 and give the angle theta at the centre — Source: Circles & chords by Save My Exams

So θ = 360 − 90 − 90 − 25
θ = 155°
Check: 155 + 90 + 90 + 25 = 360

5. Symmetry properties of a circle

Three further facts

Equal chords are equidistant from the centre, and the statement works in reverse too, so chords equidistant from the centre have equal length
The perpendicular bisector of a chord passes through the centre, which means a radius drawn to the midpoint of a chord meets it at a right angle and cuts it exactly in half

A circle with centre O and a chord across it, with a line drawn from O to the chord meeting it at a right angle and the two halves of the chord marked equal by matching dashes — Source: Circles & chords by Save My Exams

Tangents from an external point are equal in length, which is the syllabus's own wording, and the two tangents with the two radii form a kite that has a line of symmetry
The chord fact creates a right-angled triangle from the radius, half the chord and the distance to the centre, which is where Pythagoras comes in

A chord PQ with its midpoint M joined to the centre O, the right angle at M marked, PM and MQ marked equal, and the radius OQ of 6 cm and the 40 degree angle at Q shown, so triangle OMQ can be solved — Source: Circles & chords by Save My Exams

The tangent fact creates two congruent right-angled triangles back to back

Worked example

Pythagoras on a bisected chord

A circle has centre O and radius 13 cm. A chord PQ is 24 cm long. Find the distance from O to the chord.

A circle with centre O. A horizontal chord PQ below the centre is labelled 24 cm, and the radius OP is labelled 13 cm. A dashed line runs from O down to the point M on the chord, meeting it at a marked right angle. The length of that dashed line is not labelled.

Solution:

Let M be the midpoint of PQ, and join OM
The perpendicular bisector of a chord passes through the centre, so OM meets PQ at 90° and PM = MQ
So PM = 24 ÷ 2 = 12 cm
Triangle OMP has a right angle at M, with OP = 13 cm as the hypotenuse
By Pythagoras, OM² = 13² − 12² = 169 − 144 = 25
OM = √25 = 5 cm
Check: 5² + 12² = 25 + 144 = 169 = 13²