Picture a circle with centre O, and two points P and Q on the circumference. Draw the two radii OP and OQ — that gives you the "angle at the centre." Now pick any third point on the circumference (on the major arc) and join it to both P and Q — that gives you the "angle at the circumference." As long as both angles are subtended by the same arc PQ, the centre angle is always exactly double the circumference angle.

How to spot it: find two radii going to the circumference, then check if there's a third point connected to those same two endpoints. It often looks like an arrowhead pointing into the circle — but it's still true even when the triangle "parts" overlap, or when the lines form a diamond shape (in which case you compare the reflex angle at the centre with the circumference angle).

Worked example: In a circle with centre O, angle OAB = angle OBA = 60° (base angles of an isosceles triangle, since OA = OB = radius), and the angle at the centre (reflex, made of two parts) is 150°. If the angle at the circumference is (x + 60)°, then using the theorem:

This is actually just the centre/circumference theorem in disguise! If P and Q are the two ends of a diameter, then the "angle at the centre" between them is a straight line — 180°. Half of 180° is 90°. So any point on the circumference, joined to both ends of a diameter, always forms a right angle.

Worked example: P, Q, R are points on a circle, and RQ is a diameter. Angle RQP (at Q) = 40°, and the angle at P (angle QPR) = 90° because it's the angle in a semicircle. To find angle y at R:

Sometimes these questions ask you to find a length, not an angle — using the right angle you now know exists, you can bring in Pythagoras' Theorem or SOHCAHTOA on the triangle.

A chord is simply a straight line joining any two points on the circumference (it doesn't have to pass through the centre — that's a diameter's job). Here's the neat fact: if you draw a line from the centre that hits the midpoint of a chord, that line will always cut the chord at a perfect right angle. This creates two congruent (identical) right-angled triangles, each with a line of symmetry.

Worked example: Circle centre O, radius 6 cm. P and Q are on the circumference, angle OQP = 40°. Draw OM to the midpoint M of PQ — this creates a right angle at M. Using SOHCAHTOA on triangle OMQ:

A tangent is a line that touches the circle at exactly one point, without crossing into it. Wherever a tangent touches the circle, the radius drawn to that exact point is always perpendicular to the tangent — no exceptions.

If you stand outside a circle and draw two tangent lines that both touch the circle, those two tangent segments (from your point to each point of contact) are always exactly the same length. Join everything up and you get a kite: two tangents, two radii, with a line of symmetry down the middle and two right angles where the tangents meet the radii.

Worked example: ST and RT are tangents from external point T, touching the circle at S and R. Angle at T (angle STR) = 25°. Since a radius and tangent meet at 90°, angle TSO = angle TRO = 90°. The quadrilateral OSTR (O = centre) has angles summing to 360°:

A cyclic quadrilateral is a four-sided shape where all four corners touch the circle's circumference. For any such shape, the two pairs of opposite angles each add up to 180°.

Crucial condition: this ONLY works if all four vertices are genuinely on the circumference. A shape with one vertex at the centre (like a kite formed by two radii and two tangent points) is not a cyclic quadrilateral, even if it looks similar — don't apply this rule there.

Worked example: Circle centre O, cyclic quadrilateral with one angle (2x+4)°. A radius bisects a chord, creating two congruent triangles with base angles 72° and 72°, and using angles in a triangle, another angle of 18°. So one full vertex angle of the cyclic quadrilateral = 20 + 18 = 38°(built from two parts). Using the opposite-angles rule:

Take two fixed points P and Q on the circumference. Now pick any point on the same arc (same side of chord PQ) and join it to both P and Q. No matter where on that arc you pick your point, the angle you get is always the same. This is because all these angles are "looking at" the same arc PQ from the same side — the chord PQ splits the circle into two segments, and this rule holds within each segment.

Worked example: Circle centre O with points A, B, C, D, E. CE is a diameter, so angle EAC = 90° and angle CDE = 90° (angle in a semicircle, twice!). Working through the triangles gives angle ECA = 64° and angle ECD = 17°. Now, angle θ (angle EBD) and angle ECD are both formed from the chord ED, viewed from the same segment:

This is the trickiest one to spot, so let's build it up carefully. You need: a cyclic triangle (three points on the circumference, joined up), and a tangent touching the circle at one of those three vertices. The tangent creates an angle with one side of the triangle at that vertex. The theorem says: that angle equals the angle inside the triangle, at the far corner — specifically, the corner opposite the side that formed the tangent angle.

"Alternate segment" means: the chord (one side of the triangle) splits the circle into two segments. The angle between the tangent and the chord, on one side, equals the angle inscribed in the segment on the other (alternate) side.

How to spot it: look for a triangle with all three corners on the circle, where one corner also touches a tangent line. Mark the angle between the tangent and the nearest side of the triangle — the equal angle is the one inside the triangle, at the opposite corner from that side.

Worked example: A, B, C are points on a circle, DAC is a straight line, EBF is a tangent touching at B. Angle FBC = 47° (between tangent and chord BC). By the alternate segment theorem, angle CBF = angle CAB, so angle CAB = 47°. Since x° and angle CAB form a straight line along DAC: