What you'll learn
Circle theorems are a rich and rewarding part of the CSEC Mathematics syllabus under Geometry. They are a set of rules describing the relationships between angles, chords, tangents and arcs in a circle. Once you know them, problems that look complicated become a chain of short, logical steps. In this guide you will learn each of the main theorems CSEC expects, how to spot which one applies, and how to write the reasons examiners want to see. You will also learn to combine theorems with earlier geometry — angles on a straight line, angles in a triangle, and isosceles triangles formed by radii. Circle theorem questions appear in Paper 2 and reward students who reason clearly and quote the correct theorem at each step.
Key terms and definitions
Centre — the fixed point from which every point on the circle is equidistant.
Radius — a straight line from the centre to the circumference; all radii of a circle are equal.
Chord — a straight line joining two points on the circumference.
Diameter — a chord passing through the centre; the longest chord.
Tangent — a straight line that touches the circle at exactly one point.
Arc — part of the circumference; a minor arc is the shorter one, a major arc the longer.
Subtended angle — the angle formed at a point by two lines drawn from the ends of a chord or arc.
Cyclic quadrilateral — a four-sided figure with all four vertices on the circle.
Core concepts
Angle at the centre
The angle subtended by an arc at the centre of a circle is twice the angle subtended by the same arc at any point on the circumference. If an arc subtends 50° at the circumference, it subtends 100° at the centre. This is often the key that unlocks a problem.
Angle in a semicircle
A special case of the centre theorem: the angle in a semicircle is a right angle. If a triangle is drawn with the diameter as one side and the third vertex on the circle, the angle at that vertex is 90°. Spotting a diameter in the diagram is the clue.
Angles in the same segment
Angles subtended by the same arc (in the same segment) are equal. Two angles standing on the same chord, both on the same side, must be the same size. This produces neat, quick deductions.
Cyclic quadrilateral
The opposite angles of a cyclic quadrilateral add up to 180° (they are supplementary). So if one angle is 85°, the angle opposite it is 95°. A linked result: the exterior angle of a cyclic quadrilateral equals the interior opposite angle.
Tangent properties
A tangent meets a radius at the point of contact at 90° — the radius is perpendicular to the tangent. Also, the two tangents drawn from an external point are equal in length, and they make equal angles with the line joining that point to the centre, creating a symmetrical kite shape.
Alternate segment theorem
The angle between a tangent and a chord equals the angle subtended by that chord in the alternate segment (the segment on the other side of the chord). This is the most sophisticated CSEC circle theorem and is worth practising until you can spot it instantly.
Using radii and isosceles triangles
Because all radii are equal, any triangle formed by two radii is isosceles, so its base angles are equal. Combining this with the theorems above lets you find angles that are not directly given.
Worked examples
Example 1: Angle at the centre (Paper 2 style)
An arc subtends an angle of 110° at the centre O of a circle. What angle does it subtend at a point P on the major arc?
The angle at the centre is twice the angle at the circumference on the major arc. So the angle at P = 110° ÷ 2 = 55° (angle at centre is twice angle at circumference).
Example 2: Cyclic quadrilateral (Paper 2 style)
ABCD is a cyclic quadrilateral with angle A = 78° and angle B = 95°. Find angles C and D.
Opposite angles of a cyclic quadrilateral sum to 180°. So angle C = 180° − 78° = 102° (A and C are opposite), and angle D = 180° − 95° = 85° (B and D are opposite).
Example 3: Tangent and radius with isosceles triangle (Paper 2 style)
From an external point T, a tangent touches a circle of centre O at point A. OT = 13 cm and the radius OA = 5 cm. Find the length TA.
The radius meets the tangent at 90°, so triangle OAT is right-angled at A. By Pythagoras, TA² = OT² − OA² = 13² − 5² = 169 − 25 = 144, so TA = √144 = 12 cm (tangent ⊥ radius at point of contact).
Common mistakes and how to avoid them
Halving the wrong way. The centre angle is the larger one (double the circumference angle). If you find the circumference angle, multiply by 2 for the centre; if you have the centre angle, divide by 2.
Missing the diameter clue. Whenever a triangle uses the diameter, the opposite angle is 90° — look for it before trying anything harder.
Confusing "same segment" with "opposite segment". Equal angles must stand on the same chord and be on the same side. Opposite angles (cyclic quadrilateral) are supplementary, not equal.
Forgetting to quote reasons. CSEC awards marks for the reason, not just the number. Always write the theorem name (e.g. "angles in the same segment are equal").
Overlooking equal radii. Two radii always make an isosceles triangle; use the equal base angles.
Exam technique for Circle Theorems
Mark the diagram as you go. Write each angle you find directly on the figure; later steps usually depend on earlier ones.
Always give the reason. Each statement should be followed by the theorem you used, in words. This is where marks are won or lost.
Look for the "trigger" features. A diameter → right angle in a semicircle; a tangent → perpendicular radius or alternate segment; four points on the circle → cyclic quadrilateral.
Work in small steps. Find one angle, state the reason, then use it to find the next. Do not try to jump to the answer.
Check angle sums. Angles in a triangle sum to 180°, and angles on a straight line sum to 180° — use these to verify.
Quick revision summary
Circle theorems describe how angles, chords and tangents relate. The angle at the centre is twice the angle at the circumference standing on the same arc; the angle in a semicircle is 90°; angles in the same segment are equal; and opposite angles of a cyclic quadrilateral sum to 180°. A tangent is perpendicular to the radius at the point of contact, two tangents from an external point are equal, and the alternate segment theorem says the angle between a tangent and a chord equals the angle in the alternate segment. Because all radii are equal, radius-radius triangles are isosceles with equal base angles. To solve a problem, mark the diagram, spot the trigger feature (diameter, tangent, four concyclic points), work in small logical steps, and — crucially for CSEC marks — quote the correct theorem as the reason for every angle you find. Verify your result using the triangle and straight-line angle sums.