Circle Theorems: IGCSE Maths
Circle theorems are a set of rules about the angles and lengths formed inside a circle, and they appear on every Edexcel IGCSE Maths paper. Once you learn to recognise the signature shape of each theorem, these questions become some of the most rewarding marks available. This page walks through the six core circle theorems with clear diagrams, worked examples and free auto-marked practice.
Parts of a Circle
- Centre
- The fixed point (usually labelled \(O\)) that every point on the circle is the same distance from.
- Radius
- A straight line from the centre to the circumference. Every radius is the same length.
- Diameter
- A chord that passes through the centre. Its length is twice the radius.
- Chord
- A straight line joining any two points on the circumference.
- Tangent
- A straight line that touches the circle at exactly one point, called the point of contact.
- Arc
- A curved section of the circumference between two points.
- Segment
- The region between a chord and the arc it cuts off.
The Six Circle Theorems
Every angle question on a circle diagram is solved by spotting which of these six rules applies. Each theorem has a signature shape. Train your eye to recognise the shape and the rule follows.
Exam language matters. Examiners award marks for giving the correct reason, not just the correct answer. Vague phrases like "circle rule", "angle at origin is twice the angle at the edge", or "angle between tangent and circle is 90°" have been explicitly marked as not accurate enough in Edexcel examiner reports. Learn the exact wording below word-for-word, and always write the full sentence out in your working.
Worked Examples
Each example below uses one of the six theorems. The final combined example mixes two theorems in a single problem.
💡 Example 1: Angle at the Centre
Points \(A\) and \(B\) lie on a circle with centre \(O\). Point \(P\) lies on the major arc. The angle \(AOB = 128^\circ\). Find the angle \(APB\).
Angle at centre \(= 2 \times\) angle at circumference.
\[ \angle AOB = 2 \times \angle APB \]
\[ 128 = 2 \times \angle APB \]
\[ \angle APB = \frac{128}{2} = 64^\circ \]
What's happening?
Both angles sit on the same chord \(AB\). \(AOB\) is at the centre, \(APB\) is at the circumference, so the one at the centre is twice the one at the circumference. Halve to reverse it.
💡 Example 2: Angles in the Same Segment
Points \(P\) and \(Q\) lie on the major arc of a circle, both looking down at the chord \(AB\). The angle \(APB = 42^\circ\). Find the angle \(AQB\).
Angles in the same segment, subtended by the same chord, are equal.
\[ \angle APB = \angle AQB \]
\[ \angle AQB = 42^\circ \]
What's happening?
\(P\) and \(Q\) are both on the same side of the chord \(AB\), and both angles open onto that chord. The rule says they must be equal, no calculation needed.
💡 Example 3: Angle in a Semicircle
\(AB\) is a diameter of a circle. Point \(C\) lies on the circumference. Triangle \(ABC\) has the angle \(BAC = 34^\circ\). Find the other two angles of the triangle.
The angle at \(C\) is in a semicircle, so:
\[ \angle ACB = 90^\circ \]
Angles in a triangle sum to \(180^\circ\):
\[ \begin{array}{rcl} \angle ABC &=& 180 - 90 - 34 \\ &=& 56^\circ \end{array} \]
What's happening?
Because \(AB\) is a diameter, the angle at \(C\) is guaranteed to be \(90^\circ\). Once you have one angle and a right angle, standard triangle rules finish the job.
💡 Example 4: Cyclic Quadrilateral
\(ABCD\) is a quadrilateral with all four vertices on a circle. The angle \(DAB = 73^\circ\). Find the angle \(BCD\).
Opposite angles of a cyclic quadrilateral sum to \(180^\circ\).
\[ \angle DAB + \angle BCD = 180 \]
\[ 73 + \angle BCD = 180 \]
\[ \angle BCD = 180 - 73 = 107^\circ \]
What's happening?
\(A\) and \(C\) are opposite corners of the quadrilateral (they are not joined by a side), so their angles must add to \(180^\circ\). Subtract the known angle.
💡 Example 5: Tangent Meets Radius
A tangent touches a circle at point \(T\). The centre is \(O\), and \(P\) is a point on the tangent such that \(OP = 25\) cm and the radius \(OT = 7\) cm. Find the length \(PT\).
The tangent meets the radius at \(T\) at \(90^\circ\), so triangle \(OTP\) is right-angled at \(T\).
By Pythagoras:
\[ OP^2 = OT^2 + PT^2 \]
\[ 25^2 = 7^2 + PT^2 \]
\[ 625 = 49 + PT^2 \]
\[ PT^2 = 576 \]
\[ PT = 24 \text{ cm} \]
What's happening?
The tangent-radius theorem gives us a right angle at \(T\) for free. A right-angled triangle is hidden inside the diagram. Once you spot it, Pythagoras does the rest.
💡 Example 6: Two Tangents
Two tangents are drawn from an external point \(P\) to a circle, touching the circle at points \(A\) and \(B\). \(PA = 3x + 2\) and \(PB = 5x - 6\). Find the value of \(x\) and the length of each tangent.
Tangents from an external point are equal in length.
\[ PA = PB \]
\[ 3x + 2 = 5x - 6 \]
\[ 8 = 2x \]
\[ x = 4 \]
Substitute back: \( PA = 3(4) + 2 = 14 \).
Length of each tangent \(= 14\).
What's happening?
Setting the two tangent lengths equal turns a circle problem into a simple linear equation. Solve for \(x\), then substitute back to find the actual length.
💡 Example 7: Combined (Two Theorems)
A circle has centre \(O\). Points \(A\), \(B\) and \(C\) lie on the circumference. \(AB\) is a diameter. The angle \(BOC = 110^\circ\). Find the angle \(BAC\).
Step 1: \(BOC\) is at the centre and \(BAC\) is at the circumference, both standing on arc \(BC\).
Angle at centre \(= 2 \times\) angle at circumference.
\[ \angle BOC = 2 \times \angle BAC \]
\[ 110 = 2 \times \angle BAC \]
\[ \angle BAC = 55^\circ \]
Check: \(AB\) is a diameter, so the angle \(ACB = 90^\circ\) (angle in semicircle). Then \(\angle ABC = 180 - 90 - 55 = 35^\circ\). The three angles of triangle \(ABC\) sum to \(180^\circ\), which confirms the answer.
What's happening?
Two theorems live in one diagram. The angle at the centre theorem gives the unknown directly; the semicircle theorem gives a free check. In an exam, using a second theorem to verify is a strong habit that catches errors.
🔑 Key Points
- Angle at centre is twice the angle at the circumference (same arc).
- Angles in the same segment, from the same chord, are equal.
- An angle in a semicircle is always \(90^\circ\).
- Opposite angles in a cyclic quadrilateral sum to \(180^\circ\).
- The angle between a tangent and a radius is \(90^\circ\).
- Tangents drawn from an external point are equal in length.
⚠️ Pitfalls
- The "angle at centre" theorem only works when both angles stand on the same arc. Check which side of the chord each angle sits on.
- Adjacent angles in a cyclic quadrilateral do not add to \(180^\circ\). Only opposite angles do.
- Do not assume a triangle is isosceles unless two sides are radii (both equal to \(r\)). A triangle formed by two chords and a radius is usually not isosceles.
- "Angle in a semicircle" requires the chord to be a diameter, not just any chord through the circle.
- Vague or loose wording loses the reason mark. "Circle rule", "it's half", "angle at origin", or "angle between tangent and circle" are all marked as not accurate enough in Edexcel examiner reports. Always write the full theorem.