M33 · Circle theorems | Edexcel GCSE Maths

Revise the key ideas

Use circle theorems only when their conditions hold: identify the centre, the chord or arc involved, and points actually on the circumference. Give the theorem as the reason.
The angle at the centre is twice the angle at the circumference standing on the same arc. An angle of 50° at the circumference corresponds to 100° at the centre. Choose the correct arc and centre angle.
Centre angle twice circumference angle
Enlarge diagram: Centre angle twice circumference angle
Angles in the same segment are equal: two circumference angles subtended by the same chord from the same side of it are equal.
An angle in a semicircle is 90°. A triangle with one side as a diameter and its other vertex on the circle is right-angled.
Worked example: If AB is a diameter and C lies on the circumference, ∠ACB = 90°. If ∠CAB = 32°, then ∠ABC = 58° by the triangle angle sum.
A right angle subtended by a diameter
Enlarge diagram: A right angle subtended by a diameter
Opposite angles of a cyclic quadrilateral sum to 180°. All four vertices must lie on the circle; an arbitrary quadrilateral does not qualify.
Opposite angles of a cyclic quadrilateral
Enlarge diagram: Opposite angles of a cyclic quadrilateral

A tangent is perpendicular to the radius at the contact point. Mark the 90° angle before using triangle-angle rules.
A radius meeting a tangent at a right angle
Enlarge diagram: A radius meeting a tangent at a right angle
Tangents drawn from the same external point have equal lengths. If PA and PB touch the circle at A and B, PA = PB.
The perpendicular from the centre to a chord bisects the chord. If a radius and half-chord form a right triangle, Pythagoras can find their distances.
The alternate segment theorem says the angle between a tangent and a chord equals the angle subtended by that chord in the opposite segment. Match the chord endpoints carefully.
Alternate segment theorem
Enlarge diagram: Alternate segment theorem

Worked example: A cyclic quadrilateral has angle A = 112°. Its opposite angle C = 180° − 112° = 68°, by the cyclic quadrilateral theorem.
Worked example: Two tangents from P meet at 60°. Joining the centre O to both contacts creates two right angles. The angle between the radii is 360° − 90° − 90° − 60° = 120°.
To prove the centre-angle theorem in a suitable configuration, join radii to create isosceles triangles, use equal base angles and angle sums, then express the centre angle in terms of the circumference angle.
For a proof with centre O inside triangle ACB, let ∠ACO = p and ∠OCB = q. Equal radii give ∠AOC = 180° − 2p and ∠BOC = 180° − 2q. Angles around O give the remaining angle AOB = 2(p + q) = 2∠ACB. Other placements require the corresponding angle subtraction.
For equal tangents, triangles OAP and OBP are right-angled, have common hypotenuse OP and equal radii OA = OB. RHS congruence gives PA = PB.
Do not rely on how an angle looks in a drawing. Use algebraic angle labels and a sequence of valid theorems for a proof; different placements may require adding rather than subtracting angles.

Practice questions with explained answers will be added here.

For now, cover the worked answers, try the calculations yourself, then compare each step. Include units and reasons where needed.