Applying Circle Geometry Facts Revision Notes | GCSE AQA Higher Further Maths

1. Four Facts That Unlock Circle Problems

Circle geometry questions on GCSE papers often look complicated at first glance. The secret is that nearly every one of them is solved by applying one of just four facts. Once you know which fact fits the diagram, the algebra becomes straightforward.

These four facts are not arbitrary rules to memorise in isolation — each one follows from the symmetry of the circle. Recognising the right fact quickly is the skill the examiner is testing.

2. Angle in a Semi-Circle

Picture a triangle drawn inside a circle so that one of its sides is the diameter. The angle at the circumference — the vertex touching the arc — is always exactly $90^\circ$. This holds wherever that vertex sits on the arc.

The key trigger in a question is spotting that a triangle has its longest side as the diameter. Once you see that, write $\angle APB = 90^\circ$ and use it immediately — perhaps to find another angle using the angle sum in a triangle ($180^\circ$).

Worked example: Using the Angle in a Semi-Circle

A, B, and P are points on a circle. AB is a diameter. Angle PAB = 38°. Find angle APB and then angle ABP.

1. $AB$ is a diameter, so $\angle APB = 90^\circ$ — Angle in a semi-circle theorem — the angle at the circumference on a diameter is always 90°. [M1]
2. $\angle APB + \angle PAB + \angle ABP = 180^\circ$ — Angles in any triangle sum to 180°.
3. $90^\circ + 38^\circ + \angle ABP = 180^\circ$ — Replace the two known angles.
4. $\angle ABP = 180^\circ - 90^\circ - 38^\circ = 52^\circ$ — $180 - 90 - 38 = 52$. [A1]

Answer: $\angle APB = 90^\circ$ and $\angle ABP = 52^\circ$

3. Perpendicular from the Centre to a Chord

When a line from the centre of a circle meets a chord at right angles, it cuts that chord into two equal halves. This is the perpendicular bisector property of chords, and it works in both directions: if you know a line from the centre bisects a chord, that line must be perpendicular to the chord.

The most common exam application is finding a missing length using Pythagoras' theorem. You have a right-angled triangle formed by the radius, the perpendicular distance from the centre to the chord, and half the chord length.

Worked example: Finding a Chord Length Using the Perpendicular from the Centre

A circle has centre O and radius 13 cm. A chord AB is 10 cm from the centre (measured perpendicularly). Find the length of AB.

1. Let $M$ be the midpoint of $AB$. Then $OM \perp AB$, so triangle $OMA$ is right-angled at $M$. — The perpendicular from the centre bisects the chord, so $M$ is the midpoint of $AB$. [M1]
2. $OA = 13$ cm (radius), $OM = 10$ cm (perpendicular distance) — These are the hypotenuse and one leg of the right-angled triangle.
3. $AM^2 = OA^2 - OM^2 = 13^2 - 10^2$ — $13^2 = 13 \times 13 = 169$ and $10^2 = 10 \times 10 = 100$.
4. $AM^2 = 169 - 100 = 69$, so $AM = \sqrt{69}$ cm — $\sqrt{69} \approx 8.307$ cm.
5. $AB = 2 \times AM = 2\sqrt{69} \approx 16.6$ cm — Because $M$ is the midpoint, the full chord is twice $AM$. [A1]

Answer: $AB = 2\sqrt{69} \approx 16.6$ cm

4. The Tangent-Radius Angle

A tangent touches the circle at exactly one point. At that point, the radius and the tangent are perpendicular — they meet at exactly $90^\circ$. This right angle is your entry point for using Pythagoras or trigonometry in any tangent problem.

Worked example: Finding a Length Using the Tangent-Radius Property

O is the centre of a circle with radius 5 cm. PT is a tangent to the circle at T, and OP = 13 cm. Find the length PT.

1. $\angle OTP = 90^\circ$ because the tangent is perpendicular to the radius at $T$. — Tangent-radius theorem. This gives us a right-angled triangle OTP. [M1]
2. Hypotenuse $OP = 13$ cm, one leg $OT = 5$ cm (radius), find $PT$. — OP is the hypotenuse because it is opposite the right angle.
3. $PT^2 = OP^2 - OT^2 = 13^2 - 5^2$ — $13^2 = 13 \times 13 = 169$ and $5^2 = 5 \times 5 = 25$.
4. $PT^2 = 169 - 25 = 144$, so $PT = \sqrt{144} = 12$ cm — $\sqrt{144} = 12$ exactly. [A1]

Answer: $PT = 12$ cm

5. Equal Tangents from an External Point

Draw two tangent lines from the same point outside a circle. No matter where that external point is, the two tangent lengths — measured from the external point to each point of tangency — are always equal.

This fact is particularly useful when a question gives you one tangent length and asks for another, or when it sets up an equation involving two tangent expressions. Spotting that the two lengths are equal is the key step.

Worked example: Using Equal Tangents from an External Point

P is a point outside a circle with centre O. PA and PB are tangents to the circle, where A and B are the points of tangency. PA = (3x + 2) cm and PB = (5x - 6) cm. Find the length of each tangent.

1. $PA = PB$ because tangents from an external point are equal in length. — Equal tangents theorem. [M1]
2. $3x + 2 = 5x - 6$ — Replace PA and PB with their expressions.
3. $2 + 6 = 5x - 3x \Rightarrow 8 = 2x \Rightarrow x = 4$ — Collect $x$ terms on one side and constants on the other. [M1]
4. $PA = 3(4) + 2 = 12 + 2 = 14$ cm — Substitute $x = 4$ into either expression. Check: $PB = 5(4) - 6 = 20 - 6 = 14$ cm. Both match. [A1]

Answer: $PA = PB = 14$ cm

6. Putting It All Together — Practice

Exam questions often combine two of these facts in a single diagram. The strategy is always the same: identify which fact applies to each part of the diagram, write down the angle or length it gives you, then use that result in the next step.

7. Exam Technique and Common Mistakes

Circle geometry questions reward students who state their reasoning clearly. Writing the name of the theorem — for example, 'angle in a semi-circle' or 'tangent-radius angle' — alongside the value earns the method mark even if a later arithmetic slip costs the accuracy mark.

The most reliable approach on any circle question is to annotate the diagram first. Mark right angles where you know them, label equal lengths, and write in any angle values you can deduce before writing a single line of algebra.