Circle Geometry is worth ~40 marks in Paper 2 and is one of the topics students fear most. But every single question uses the same small set of theorems. If you understand the logic behind each one, you can solve ANY circle geometry problem.
The golden rule of geometry proofs: Every statement needs a reason. No reason = no mark. Learn the exact wording for each theorem.
How to Approach ANY Circle Geometry Problem#
- Mark what you know — fill in given angles, equal sides, right angles.
- Look for the patterns — which theorem does this look like?
- Write Statement → Reason for every single line.
- Check your angle sum — angles in a triangle = $180°$, angles on a straight line = $180°$.
Theorem 1: Perpendicular from Centre to Chord#
A line drawn from the centre of a circle perpendicular to a chord bisects (halves) the chord.
Reason to write: line from centre $\perp$ to chord
Converse: A line from the centre to the midpoint of a chord is perpendicular to it.
Reason: line from centre to midpt of chord
Why it works#
Draw radii to both endpoints of the chord. You create two right-angled triangles that are congruent (RHS: same hypotenuse = radius, same right angle, shared side). So the chord must be split equally.
Worked Example#
O is the centre. Chord AB = 24 cm. The perpendicular distance from O to AB is 5 cm. Find the radius.
| Statement | Reason |
|---|---|
| Let M be the foot of the perpendicular from O to AB | Construction |
| $AM = MB = 12$ cm | line from centre $\perp$ to chord |
| $OM = 5$ cm | Given |
| $OA^2 = OM^2 + AM^2 = 25 + 144 = 169$ | Pythagoras |
| $OA = 13$ cm |
The radius is 13 cm.
How to spot it#
Look for a line from the centre hitting a chord. If it’s perpendicular, the chord is halved. If it hits the midpoint, it’s perpendicular.
Theorem 2: Angle at the Centre = 2 × Angle at Circumference#
The angle subtended by an arc at the centre is double the angle subtended by the same arc at the circumference.
Reason: $\angle$ at centre = $2 \times \angle$ at circumf
Why it works#
Draw a radius from O to the vertex on the circumference. This creates two isoceles triangles (two sides are radii). The base angles of each are equal. When you add them up, the centre angle is always double.
Worked Example#
O is the centre. $\hat{O} = 140°$. Find the angle at the circumference.
| Statement | Reason |
|---|---|
| Let $\hat{C}$ be the angle at the circumference, standing on the same arc | |
| $\hat{O} = 2\hat{C}$ | $\angle$ at centre = $2 \times \angle$ at circumf |
| $140° = 2\hat{C}$ | |
| $\hat{C} = 70°$ |
⚠️ Watch out#
Both angles must stand on the same arc. If the angle at the circumference is on the wrong side (the major arc instead of the minor arc, or vice versa), this theorem doesn’t apply directly.
How to spot it#
Look for a “kite” or “arrowhead” shape — two lines from an arc meeting at the centre, and two lines from the same arc meeting at the circumference.
Theorem 3: Angle in a Semicircle = 90°#
The angle subtended by a diameter at the circumference is always $90°$.
Reason: $\angle$ in semi-circle
Why it works#
This is a special case of Theorem 2. The diameter creates a straight angle ($180°$) at the centre. Half of $180°$ is $90°$.
Worked Example#
AB is a diameter. C is on the circle. $\hat{A} = 35°$. Find $\hat{B}$.
| Statement | Reason |
|---|---|
| $\hat{C} = 90°$ | $\angle$ in semi-circle (AB is diameter) |
| $\hat{A} + \hat{B} + \hat{C} = 180°$ | $\angle$ sum of $\triangle$ |
| $35° + \hat{B} + 90° = 180°$ | |
| $\hat{B} = 55°$ |
How to spot it#
Whenever you see a diameter (a chord passing through the centre), any angle touching the circumference from that diameter is $90°$.
Theorem 4: Angles in the Same Segment#
Angles subtended by the same arc (or chord) at the circumference, on the same side of the chord, are equal.
Reason: $\angle$s in same segment
Why it works#
Both angles equal half the centre angle (Theorem 2). Half of the same thing = same answer.
Worked Example#
A, B, C, D are on a circle. $\hat{C} = 42°$ and both $\hat{C}$ and $\hat{D}$ are subtended by chord AB on the same side. Find $\hat{D}$.
| Statement | Reason |
|---|---|
| $\hat{D} = \hat{C} = 42°$ | $\angle$s in same segment (both subtended by arc AB) |
How to spot it#
Look for the “bowtie” pattern — two triangles sharing the same chord as their base, with their apex vertices on the same side of the circle.
Theorem 5: Equal Chords, Equal Angles#
Equal chords subtend equal angles at the centre (and at the circumference).
Reason: equal chords, equal $\angle$s
Converse#
Equal angles at the centre stand on equal chords.
Putting It All Together: Multi-Step Proof#
Worked Example#
O is the centre. A, B, C are on the circle. $O\hat{A}B = 25°$.
Find $A\hat{C}B$.
| Statement | Reason |
|---|---|
| $OA = OB$ (radii) | radii of circle |
| $O\hat{B}A = O\hat{A}B = 25°$ | base $\angle$s of isos $\triangle$, $OA = OB$ |
| $A\hat{O}B = 180° - 25° - 25° = 130°$ | $\angle$ sum of $\triangle$ |
| $A\hat{C}B = \frac{1}{2} \times A\hat{O}B = 65°$ | $\angle$ at centre = $2 \times \angle$ at circumf |
Another Multi-Step Example#
O is the centre. $A\hat{O}B = 104°$. C is on the major arc. D is on the minor arc. Find $A\hat{C}B$ and $A\hat{D}B$.
| Statement | Reason |
|---|---|
| $A\hat{C}B = \frac{104°}{2} = 52°$ | $\angle$ at centre = $2 \times \angle$ at circumf (same arc) |
| Reflex $A\hat{O}B = 360° - 104° = 256°$ | $\angle$s around a point |
| $A\hat{D}B = \frac{256°}{2} = 128°$ | $\angle$ at centre = $2 \times \angle$ at circumf (major arc) |
Check: $52° + 128° = 180°$ ✓ (This makes sense because ACBD is a cyclic quad — opposite angles are supplementary!)
The Complete “Reason Bank” for Core Theorems#
Memorise these exact phrases — they are what markers look for:
| Theorem | Reason to write |
|---|---|
| Perpendicular bisects chord | line from centre $\perp$ to chord |
| Midpoint implies perpendicular | line from centre to midpt of chord |
| Centre angle = 2× circumference | $\angle$ at centre = $2 \times \angle$ at circumf |
| Diameter gives 90° | $\angle$ in semi-circle |
| Same segment gives equal angles | $\angle$s in same segment |
| Isoceles triangle (radii) | radii of circle / base $\angle$s of isos $\triangle$ |
| Angles in a triangle | $\angle$ sum of $\triangle$ |
| Angles on a straight line | $\angle$s on a str line |
🚨 Common Mistakes#
- Not stating that OA and OB are radii: Before using isoceles triangle properties, you MUST first state that two sides are radii. Without this, the marker won’t give marks for the base angles being equal.
- Confusing which arc the angle stands on: The “angle at the circumference” must stand on the SAME arc as the “angle at the centre.” Draw the arc in a different colour to be sure.
- Forgetting the reflex angle: When a point is on the minor arc, the centre angle you need is the REFLEX angle ($360° - \theta$), not the one you can see directly.
- Incomplete reasons: “Angles in a circle” is NOT a valid reason. You must name the SPECIFIC theorem.
- Not marking the diagram: Before writing anything, mark ALL equal sides (radii!), all right angles, and all given angles on the diagram. This reveals the theorems you need.
💡 Pro Tip: The “Isoceles Triangle” Radar#
Every time you see the centre O connected to two points on the circle, you have an isoceles triangle (because both lines are radii). This is the single most useful observation in circle geometry — it unlocks base angles, which then feed into every other theorem.
🔗 Related Grade 11 topics:
- Cyclic Quads, Tangents & Proofs — extends these theorems to four-sided figures and tangent lines
- Analytical Geometry: Circles & Tangents — the equation of a circle and finding tangent lines algebraically