Circle Theorems Explained with Diagrams

Circle theorems are a favourite in Matric, O Level and GCSE geometry papers because they test reasoning, not just calculation. Each theorem is a short rule about angles or lengths in a circle, and once you can recognise the shape of each one in a diagram, most questions take only a few lines. Below, every theorem comes with a word-picture of its diagram so you can sketch it yourself as you read.

Key Vocabulary

• Radius: a line from the centre \( O \) to the circle.
• Chord: a line joining two points on the circle. A diameter is a chord through the centre.
• Arc: part of the circumference. A chord splits the circle into a minor and a major arc.
• Segment: the region between a chord and its arc.
• Tangent: a line touching the circle at exactly one point.
• Cyclic quadrilateral: a four-sided shape with all four vertices on the circle.
• Subtended angle: the angle an arc or chord makes at a point.

Theorem 1: Angle at the Centre

Diagram: Points \( A \) and \( B \) on the circle, centre \( O \), and a third point \( C \) on the major arc. Draw \( OA \), \( OB \), \( CA \) and \( CB \); the shape looks like an arrowhead.

Rule: The angle at the centre is twice the angle at the circumference subtended by the same arc:

$$ \angle AOB = 2\angle ACB $$

Theorem 2: Angle in a Semicircle

Diagram: A diameter \( AB \) and any point \( C \) on the circle, joined to \( A \) and \( B \).

Rule: \( \angle ACB = 90^\circ \). This is a special case of Theorem 1, since the angle at the centre is a straight angle of \( 180^\circ \).

Theorem 3: Angles in the Same Segment

Diagram: A chord \( AB \) with two points \( C \) and \( D \) on the same side of it, each joined to \( A \) and \( B \). The lines make a "bow-tie" shape.

Rule: Angles subtended by the same arc in the same segment are equal: \( \angle ACB = \angle ADB \).

Theorem 4: Cyclic Quadrilaterals

Diagram: Four points \( A, B, C, D \) on the circle joined in order.

Rule: Opposite angles add up to \( 180^\circ \):

$$ \angle A + \angle C = 180^\circ, \qquad \angle B + \angle D = 180^\circ $$

A consequence: an exterior angle of a cyclic quadrilateral equals the interior opposite angle.

Theorem 5: Tangent and Radius

Diagram: A tangent touching the circle at \( T \), with radius \( OT \) drawn.

Rule: The tangent is perpendicular to the radius at the point of contact: \( \angle OTP = 90^\circ \) for any point \( P \) on the tangent.

Theorem 6: Two Tangents from an External Point

Diagram: A point \( P \) outside the circle with two tangents touching at \( A \) and \( B \); a kite shape \( OAPB \) forms.

Rule: The tangents are equal in length, \( PA = PB \), and \( OP \) bisects \( \angle APB \).

Theorem 7: Alternate Segment Theorem

Diagram: A tangent at \( T \), a chord \( TA \), and a point \( B \) on the circle in the opposite segment, joined to \( T \) and \( A \).

Rule: The angle between the tangent and the chord equals the angle in the alternate segment:

$$ \angle(\text{tangent}, TA) = \angle TBA $$

This is the theorem students most often fail to spot, so look for any tangent meeting a chord.

Theorem 8: Perpendicular from the Centre to a Chord

Diagram: A chord \( AB \) with a line from \( O \) meeting it at right angles at \( M \).

Rule: The perpendicular from the centre bisects the chord, so \( AM = MB \). Combined with Pythagoras, this lets you find chord lengths.

Worked Example 1: Combining Angle Theorems

Question: \( A, B, C, D \) lie on a circle with centre \( O \). the non-reflex angle \( \angle AOC = 140^\circ \), \( B \) lies on the major arc and \( D \) on the minor arc. Find \( \angle ABC \) and \( \angle ADC \).

Step 1: Angle at the centre (Theorem 1). \( B \) is on the major arc, so it subtends the same minor arc \( AC \) as the \( 140^\circ \) angle:

$$ \angle ABC = \frac{140^\circ}{2} = 70^\circ $$

Step 2: Cyclic quadrilateral (Theorem 4). \( ABCD \) is cyclic, so

$$ \angle ADC = 180^\circ - 70^\circ = 110^\circ $$

Check with Theorem 1: the reflex angle \( AOC \) is \( 360^\circ - 140^\circ = 220^\circ \), and half of that is \( 110^\circ \). Correct.

Worked Example 2: Tangent Length

Question: A circle has radius 5 cm. A point \( P \) is 13 cm from the centre \( O \). Find the length of the tangent from \( P \) to the circle.

Step 1: Identify the right angle (Theorem 5). If the tangent touches at \( T \), then \( \angle OTP = 90^\circ \), with \( OP \) as the hypotenuse.

Step 2: Pythagoras.

$$ PT^2 = OP^2 - OT^2 = 13^2 - 5^2 = 169 - 25 = 144 $$

$$ PT = 12 \text{ cm} $$

By Theorem 6, the second tangent from \( P \) is also 12 cm.

Worked Example 3: Chord Length

Question: A chord is 6 cm from the centre of a circle of radius 10 cm. Find the length of the chord.

Step 1: Use Theorem 8. The perpendicular from \( O \) meets the chord at its midpoint \( M \), with \( OM = 6 \) and \( OA = 10 \).

Step 2: Pythagoras in triangle \( OMA \).

$$ AM^2 = 10^2 - 6^2 = 100 - 36 = 64 \Rightarrow AM = 8 $$

Step 3: Double it.

$$ AB = 2 \times 8 = 16 \text{ cm} $$

Worked Example 4: Alternate Segment

Question: A tangent at \( T \) makes an angle of \( 52^\circ \) with chord \( TA \). \( B \) is on the circle in the alternate segment, and \( \angle BTA = 61^\circ \). Find \( \angle TAB \).

Step 1: Alternate segment theorem. \( \angle TBA = 52^\circ \).

Step 2: Angle sum of triangle \( TAB \).

$$ \angle TAB = 180^\circ - 52^\circ - 61^\circ = 67^\circ $$

Tips

• Mark every radius as equal and look for isosceles triangles with \( O \) as the apex.
• Write the theorem name next to each step; examiners award reasoning marks for it.
• Spot the shapes: arrowhead (centre), bow-tie (same segment), quadrilateral on the circle (cyclic), tangent plus chord (alternate segment).
• Redraw a messy diagram larger and add the radii.

Common Mistakes

• Applying the cyclic quadrilateral rule when one vertex is the centre, not on the circle.
• Halving the wrong angle in Theorem 1 when the point is on the minor arc.
• Forgetting that a tangent meets a radius at \( 90^\circ \).
• Using "same segment" when the two points lie on opposite sides of the chord.

Key Takeaways

• The angle at the centre is twice the angle at the circumference; the angle in a semicircle is \( 90^\circ \).
• Angles in the same segment are equal; opposite angles of a cyclic quadrilateral sum to \( 180^\circ \).
• A tangent is perpendicular to the radius, and tangents from one external point are equal.
• The alternate segment theorem links a tangent-chord angle to an angle in the opposite segment.
• Combine the theorems with Pythagoras and isosceles triangles for length questions.

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