Properties of Circles

Circle theorems O-Level E Maths: all 8 circle properties with worked examples

By Mr Gan · O-Level E Maths · Updated July 2026 · 11 min read

O-Level E Maths tests 8 circle properties, grouped into 4 categories: basic angle facts (angles in same segment, angle in semicircle), tangent properties (tangent ⊥ radius, tangents from an external point), cyclic quadrilateral facts (angles in opposite segments, angle at centre), and chord properties (perpendicular bisector of chord, equal chords). Each has a required abbreviation you must write in your working for full marks.

The 8 properties at a glance

Grouping these into 4 categories makes them far easier to hold in memory than a flat list of 8. Use this table as your quick-revision reference — the exact wording matches what you should write in your exam working.

Category	Property	Write in working as
1. Basic	Angles in the same segment are equal	∠s in same segment
1. Basic	Angle at centre = 2 × angle at circumference	∠ at centre = 2∠ at circumference
1. Basic	Angle in a semicircle is 90°	rt. ∠ in semicircle
2. Tangent	Radius is perpendicular to tangent	tangent ⊥ radius
2. Tangent	Tangents from an external point are equal	tangents from ext. pt.
3. Cyclic Quad	Angles in opposite segments sum to 180°	∠s in opp. segments
4. Chords	Perpendicular bisector of a chord	chord bisector
4. Chords	Equal chords are equidistant from centre	equal chords

Why this topic is worth mastering properly

Circle theorems are one of the few O-Level topics where the "answer" isn't really the number you calculate — it's the reasoning. Every angle you find must be justified with the correct theorem name, written in the exact abbreviated form examiners expect. Get the angle right but the reason wrong (or missing), and you lose marks even though your maths was correct.

The good news: there are only 8 properties to know, and most O-Level questions combine 2 or 3 of them in a single diagram. Once you can spot each pattern on sight, these questions become some of the fastest marks on the paper.

Category 1 — Basic angle properties

1Angles in the same segment are equal

Any two angles subtended by the same arc AB, from points P and Q on the same side (same segment), are equal to each other.

∠APB = ∠AQB

Write in working: "∠s in same segment"

2Angle at centre = 2 × angle at circumference

The angle formed at the centre O by two radii is exactly double the angle formed at the circumference by the same two points, as long as both angles stand on the same arc AB.

∠AOB = 2 × ∠APB

Write in working: "∠ at centre = 2∠ at circumference"

3Angle in a semicircle = 90°

When AB is a diameter (passes through the centre O), any angle subtended at the circumference from AB is exactly 90°.

∠APB = 90°

Write in working: "rt. ∠ in semicircle"

Category 2 — Tangent properties

4Tangent is perpendicular to radius

A tangent to a circle is always perpendicular to the radius drawn to the point of contact T.

OT ⊥ tangent at T

Write in working: "tangent ⊥ radius"

5Tangents from an external point are equal

If two tangents are drawn from the same external point E, touching the circle at T₁ and T₂, their lengths are equal.

ET₁ = ET₂

Write in working: "tangents from ext. pt."

Category 3 — Cyclic quadrilateral properties

6Angles in opposite segments

In a cyclic quadrilateral ABCD (all 4 vertices on the circle), angle B and angle D lie in opposite segments of chord AC, and every such opposite pair sums to 180°.

∠B + ∠D = 180°
∠A + ∠C = 180°

Write in working: "∠s in opp. segments"

Category 4 — Chord properties

7Perpendicular bisector of a chord

A line from the centre O that is perpendicular to a chord AB always bisects that chord (cuts it exactly in half at M) — and this works both ways: if a line from O bisects the chord, that line must be perpendicular to it.

OM ⊥ AB ⇔ AM = MB

Write in working: "chord bisector"

8Equal chords are equidistant from the centre

If chord AB and chord CD are equal in length, then their perpendicular distances from the centre O are also equal — and again, this works both ways.

AB = CD ⇔ dist. from O equal

Write in working: "equal chords"

Watch the whole method in about a minute.

Worked example — perpendicular bisector of a chord

A chord AB in a circle of radius 13 cm is 24 cm long. Find the perpendicular distance from the centre O to the chord.

Solution

Draw the perpendicular from O to AB, meeting it at M. By property 7, OM bisects AB, so AM = MB = 24 ÷ 2 = 12 cm.

OA is a radius, so OA = 13 cm. Triangle OMA is right-angled at M (OM ⊥ AB).

Apply Pythagoras' theorem:
OM² = OA² − AM² = 13² − 12² = 169 − 144 = 25

Square root:
OM = √25 = 5 cm

Recognise this pattern: "chord + radius + perpendicular distance" almost always means drawing the perpendicular from the centre, using property 7 to halve the chord, then applying Pythagoras in the right-angled triangle formed. This exact setup reappears constantly in mensuration and circle questions.

Worked example — combining 3 properties in one question

In the diagram, O is the centre of the circle. A, B, C, D lie on the circumference. AC is a diameter. ∠ABD = 35° and ∠BDC = 40°. Find ∠BCA and ∠BOC.

Solution

∠BDC and ∠BAC are both subtended by the same arc BC, so they're angles in the same segment:
∠BAC = ∠BDC = 40° (∠s in same segment)

Since AC is a diameter, ∠ABC is the angle in a semicircle:
∠ABC = 90° (rt. ∠ in semicircle)

∠ABC is made up of ∠ABD + ∠DBC. We know ∠ABD = 35°, so:
∠DBC = 90° − 35° = 55°

In triangle BDC, angles sum to 180°. We know ∠BDC = 40° and ∠DBC = 55°:
∠BCD = 180° − 40° − 55° = 85°

In triangle ABC, angles sum to 180°. We know ∠BAC = 40° and ∠ABC = 90°:
∠BCA = 180° − 40° − 90° = 50°

∠BOC is the angle at the centre standing on the same arc BC as ∠BAC (angle at circumference):
∠BOC = 2 × ∠BAC = 2 × 40° = 80° (∠ at centre = 2∠ at circumference)

The pattern in every multi-step circle question: find one angle using a property, then use that result (plus angle sum of triangle = 180°, or angles on a straight line = 180°) to unlock the next angle. Circle theorem questions are rarely solved with a single property — expect to chain 2 or 3 together.

The two things that actually cost marks

1. Forgetting the reason

Writing "∠BAC = 40°" without stating "(∠s in same segment)" alongside it loses the reasoning mark, even if the number is correct. Every angle derived from a circle property needs its abbreviation written next to it — this is non-negotiable in the O-Level mark scheme.

2. Misidentifying the segment or the diameter

Property 2 (angle in semicircle = 90°) only applies when the chord is actually a diameter — passing through the centre. Students sometimes apply it to any chord that merely looks long in the diagram. Always confirm the diameter is explicitly stated or marked before using this property.

Quick recognition guide, by category

Basic — two triangles sharing the same chord, both with their third vertex on the same arc? → Angles in the same segment. A chord explicitly passing through the centre (a diameter)? → Angle in a semicircle = 90°. A triangle formed by 2 radii and a chord, plus one formed by the same chord and a point on the circumference? → Angle at centre = 2 × angle at circumference.

Tangent — a straight line just touching the circle at one point? → Perpendicular to the radius there. Two such lines from the same outside point? → Equal in length.

Cyclic Quad — a 4-sided shape with all corners touching the circle? → Angles in opposite segments sum to 180°.

Chords — a line from the centre O drawn to a chord with a right angle marked? → That line bisects the chord (and vice versa). Two chords marked as equal in length? → Equal distance from the centre.

Your 8-property memorisation checklist

Category	Property	Write in working as
1. Basic	Angles in the same segment are equal	∠s in same segment
1. Basic	Angle at centre = 2 × angle at circumference	∠ at centre = 2∠ at circumference
1. Basic	Angle in a semicircle is 90°	rt. ∠ in semicircle
2. Tangent	Radius is perpendicular to tangent	tangent ⊥ radius
2. Tangent	Tangents from an external point are equal	tangents from ext. pt.
3. Cyclic Quad	Angles in opposite segments sum to 180°	∠s in opp. segments
4. Chords	Perpendicular bisector of a chord	chord bisector
4. Chords	Equal chords are equidistant from centre	equal chords

Frequently asked questions

Do I need to memorise the exact abbreviations, or can I explain in full sentences?

The standard abbreviations (e.g. "∠s in same segment", "∠s in opp. segments") are what O-Level mark schemes look for, and they're faster to write under exam time pressure. A full correct sentence explaining the same reasoning is also accepted, but the abbreviation is safer and quicker — memorise them exactly as shown in your syllabus formula list.

Why does the perpendicular bisector property work in both directions?

It's an "if and only if" relationship: a line from the centre perpendicular to a chord will always bisect it, and separately, a line from the centre that bisects a chord will always be perpendicular to it. In exam questions you might be given either piece of information — the perpendicularity or the bisection — and asked to conclude the other, so both directions are equally examinable.

What's the difference between "angle at centre" and "angle at circumference"?

The angle at centre is formed by two radii meeting at the circle's centre O. The angle at circumference is formed at any point on the circle's edge, using the same two endpoints on the arc. They must be subtended by the same arc for the ×2 relationship to hold — always check this before applying the property.

Can a cyclic quadrilateral have a reflex angle?

No — for a standard convex cyclic quadrilateral, all interior angles are between 0° and 180°, and each pair of angles in opposite segments sums to exactly 180°. If a calculated angle comes out negative or above 180°, re-check whether you've used the correct opposite pair.

How do I know which chord is a diameter if the question doesn't say so directly?

A diameter is only confirmed if the question explicitly states it (e.g. "AC is a diameter") or if the diagram clearly shows the line passing through a labelled centre point O. Never assume a chord is a diameter just because it looks long or central in a sketch — this assumption is a common source of incorrect working.

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