Circle Theorems in IGCSE Maths
All seven circle theorems with clear diagrams and worked examples. Learn the rules, spot them in diagrams, and give the correct reasons to earn full marks.
0580 · E6.2 / E6.3Angle at the centre
The angle at the centre of a circle is twice the angle at the circumference, when both are subtended by the same arc.
If the angle at the circumference (\(P\)) is \(x\), the angle at the centre (\(O\)) is \(2x\).
The angle at the centre is \(124°\). Find the angle at the circumference.
The angle at the circumference is half the angle at the centre:
\[\text{Angle} = \frac{124°}{2} = 62°\]Reason: The angle at the centre is twice the angle at the circumference.
Angle in a semicircle
The angle in a semicircle is always 90°. If a triangle is drawn inside a circle with the diameter as one side, the angle opposite the diameter is a right angle.
This is a special case of theorem 1. The angle at the centre is \(180°\) (a straight line), so the angle at the circumference is \(90°\).
Look for a triangle inside a circle where one side passes through the centre. If you see a diameter as one side of a triangle, the opposite angle is \(90°\).
Angles in the same segment
Angles subtended by the same arc at the circumference are equal.
If two angles at the circumference stand on the same chord and are on the same side of the chord, they are equal: \(\angle APB = \angle AQB\).
Points \(A\), \(B\), \(P\) and \(Q\) lie on a circle. Angle \(APB = 35°\). Find angle \(AQB\).
Reason: Angles in the same segment are equal.
Cyclic quadrilaterals
The opposite angles of a cyclic quadrilateral add up to 180°.
A cyclic quadrilateral has all four vertices on the circle. Opposite angles are supplementary:
\[a + c = 180° \qquad \text{and} \qquad b + d = 180°\]\(ABCD\) is a cyclic quadrilateral. Angle \(A = 110°\) and angle \(B = 73°\). Find angles \(C\) and \(D\).
Opposite angles sum to \(180°\):
\[\begin{aligned}\angle C &= 180° - 110° = 70° \\[6pt]\angle D &= 180° - 73° = 107°\end{aligned}\]Reason: Opposite angles of a cyclic quadrilateral are supplementary.
Tangent & radius
A tangent to a circle is perpendicular to the radius at the point of contact.
The angle between the tangent and the radius is always \(90°\). This is frequently used with Pythagoras' theorem to find missing lengths.
A circle has centre \(O\) and radius 5 cm. A tangent from an external point \(P\) touches the circle at \(T\). If \(OP = 13\) cm, find the length \(PT\).
Since the tangent is perpendicular to the radius, triangle \(OTP\) is right-angled at \(T\).
\[\begin{aligned}PT^2 &= OP^2 - OT^2 \\[6pt]&= 13^2 - 5^2 \\[6pt]&= 169 - 25 = 144 \\[6pt]PT &= 12 \text{ cm}\end{aligned}\]Reason: A tangent is perpendicular to the radius at the point of contact.
Two tangents from a point
Two tangents drawn to a circle from the same external point are equal in length.
\(PA = PB\). The two tangent lengths from any external point are always equal. This also means triangle \(OAP\) is congruent to triangle \(OBP\).
Questions often combine this with the tangent-radius theorem. Since \(OA \perp PA\) and \(OB \perp PB\), and \(PA = PB\), you can find angles in the "kite" shape \(OAPB\) using symmetry.
Alternate segment theorem
The angle between a tangent and a chord is equal to the angle in the alternate segment.
The angle between the tangent at \(T\) and chord \(TA\) equals the angle \(TBA\) in the alternate segment. This is the theorem students find hardest to recognise in exam diagrams.
A tangent at point \(T\) makes an angle of \(48°\) with chord \(TA\). Find angle \(TBA\), where \(B\) is a point on the major arc.
Reason: Alternate segment theorem — the angle between a tangent and a chord equals the angle in the alternate segment.
Exam-style question
\(A\), \(B\), \(C\) and \(D\) are points on a circle with centre \(O\). A tangent is drawn at \(C\).
Angle \(DAB = 72°\). The angle between the tangent at \(C\) and chord \(CB\) is \(54°\).
(a) Find angle \(BCD\). [2]
(b) Find angle \(BDC\). [2]
(c) Find angle \(DBC\). [2]
Show solution (a)
Opposite angles of a cyclic quadrilateral sum to \(180°\):
\[\angle BCD = 180° - 72° = 108°\]Show solution (b)
By the alternate segment theorem, the angle between the tangent at \(C\) and chord \(CB\) equals angle \(CDB\) in the alternate segment:
\[\angle BDC = 54°\]Show solution (c)
In triangle \(BCD\), angles sum to \(180°\):
\[\begin{aligned}\angle DBC &= 180° - 108° - 54° \\[6pt]&= 18°\end{aligned}\]This is typical of a 5–6 mark circle theorems question on Paper 4. It combines three theorems: cyclic quadrilateral (opposite angles), alternate segment, and angle sum in a triangle. The marks aren't just for the answers — you must name each theorem to get full marks. Always write the reason alongside your working.
Common mistakes
Getting the right angle is only half the marks. You must state which circle theorem you used. Write it in full: "Angle at the centre is twice the angle at the circumference" — not just "circle theorem".
Students sometimes double instead of halving (or vice versa). Always check: is the angle at O or at the edge? The bigger angle is at the centre.
This is the most commonly missed theorem. If you see a tangent touching the circle, always check whether there's a chord from the same point — that's the alternate segment setup.
Don't assume a line is a diameter just because it looks like one. A diameter must pass through the centre. If the diagram doesn't explicitly show this, you can't use the semicircle theorem.
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