GCSE Maths Circle theorems revision

A radius runs from centre to circumference; a diameter is the chord through the centre (and the longest chord). Any other line joining two points on the circle is a chord. A tangent brushes the circle at exactly one point. An arc is a stretch of the circumference; the region between two radii and an arc is a sector, while the region between a chord and an arc is a segment — the minor segment sits on the short side of the chord, the major segment on the long side. Exam answers lose marks for calling a segment a sector, so fix the vocabulary early.

Two drawn-in lines power almost every theorem: drawing a radius to the point of contact of a tangent creates a right angle, and drawing two radii to a chord’s ends creates an isosceles triangle. Both moves cost nothing and unlock the diagram.

The angle at the centre theorem says the angle two radii make at the centre is twice the angle the same arc subtends at the circumference: a rim angle of 36° pairs with a centre angle of 72°. Both angles must be built on the same arc — check before halving or doubling. The semicircle result follows: an arc that is a full diameter subtends 180° at the centre, so every angle it subtends at the circumference is 90°. Any triangle sitting on a diameter is right-angled at the third corner.

Angles standing on the same chord, in the same segment, are equal: two rim angles of 55° on the same side of one chord. Flip to the other side of the chord and the pair becomes cyclic: four points on a circle form a quadrilateral whose opposite angles sum to 180° — a 118° corner forces the opposite one to 62°. That is also why opposite corners of a cyclic quadrilateral always look “splayed open”.

A tangent meets the radius at 90° — draw that radius whenever a tangent appears, because the right angle starts the triangle. From an external point P, the two tangent segments to the circle have equal lengths (the line OP is a line of symmetry), and the triangle PTO is right-angled at T: with radius 5 and OP = 13, the tangent length is √(169 − 25) = 12 — a 5-12-13 in disguise. The angle at P satisfies sin P = 5⁄13, about 22.6°.

The alternate segment theorem closes the set: the angle between a tangent and a chord through the point of contact equals the angle the chord subtends in the alternate (opposite) segment. A 52° angle between tangent and chord forces every angle standing in the far segment to 52°. When proving, remember the hidden fact: a radius drawn perpendicular to a chord bisects it, so the perpendicular distance and half-chord lengths pair inside right-angled triangles — chord 16 cm at distance 6 cm from the centre gives radius √(8² + 6²) = 10 cm.

The proofs all start the same way: two radii plus a chord make an isosceles triangle, because radii are equal. Proving the centre theorem: draw the diameter through the rim point C, splitting the rim angle into x and y. Triangles OAC and OBC are isosceles, so angle OAC = x and angle OBC = y. The centre angle AOB = 2x + 2y (exterior angle of each isosceles triangle), while the rim angle ACB = x + y — double, as claimed. The semicircle result drops out when the arc is a diameter: x + y = 90°.

The alternate segment proof chains two steps: the tangent is perpendicular to the radius, so the tangent–chord angle is 90° − (base angle of the isosceles triangle); and the angle in the opposite segment is 180° − (that base angle) − (the other 90° from the diameter) — the same 90° − base. Equal by a shared computation, not by looking similar. For exam credit, name each triangle you claim is isosceles and give the radii-equal reason; the final line should restate the theorem, not trail off.