Words you need first
A circle is all the points at the same distance from a centre O. That distance is the radius.
- A chord is a straight line joining two points on the circle.
- A diameter is a chord through the centre. It is two radii long.
- An arc is a part of the edge (circumference).
- A segment is the region between a chord and an arc. A chord cuts the circle into a major (bigger) and a minor (smaller) segment.
- A tangent is a straight line that touches the circle at exactly one point.
- An angle is subtended by a chord when lines from the two ends of the chord meet at a point.
A radius drawn at right angles to a chord cuts the chord in half. So radius, half-chord and the distance from the centre form a right-angled triangle, and Pythagoras' theorem works: r² = d² + (c/2)².
Theorems about angles inside the circle
1. Angle at the centre is twice the angle at the circumference
If a chord AB makes ∠AOB at the centre and ∠APB at a point P on the edge (same side), then ∠AOB = 2 × ∠APB. The angle at the centre is called a central angle; the one on the edge is an inscribed angle.
2. Angle in a semicircle is 90°
If AB is a diameter, ∠AOB = 180°, so ∠APB = 180° ÷ 2 = 90°. This is rule 1 with a straight angle at the centre.
3. Angles in the same segment are equal
Every point on the same arc sees the chord at the same angle, because each one is half of the same central angle.
4. Opposite angles of a cyclic quadrilateral add to 180°
A cyclic quadrilateral has all four corners on one circle. ∠A + ∠C = 180° and ∠B + ∠D = 180°. Also, an exterior angle equals the interior opposite angle.
Theorems about tangents
5. Tangent and radius meet at 90°
The radius drawn to the point where a tangent touches is perpendicular to the tangent.
6. Two tangents from one outside point are equal
From a point E outside the circle, the tangent lengths ET₁ and ET₂ are equal. Triangles OT₁E and OT₂E are congruent (RHS: right angle, same hypotenuse OE, equal radii). With Pythagoras: ET = √(OE² − r²).
7. Alternate segment theorem
The angle between a tangent and a chord at the point of contact equals the inscribed angle on that chord in the alternate (other) segment.
How to prove the main theorem
Proof that the angle at the centre is double. Join P to O and extend the line. OA = OP = OB (all radii), so triangles OAP and OBP are isosceles.
- In triangle OAP, call the equal base angles x. The outside angle at O is x + x = 2x.
- In triangle OBP, call the equal base angles y. The outside angle at O is 2y.
- So ∠AOB = 2x + 2y = 2(x + y) = 2 × ∠APB. Done.
Rules 2, 3 and 4 all follow from this one. For a cyclic quadrilateral, the two central angles on chord BD add to 360°, so the two inscribed angles add to 180°.
Writing reasons in exams: always name the rule, for example "angle in a semicircle is 90°" or "opposite angles of a cyclic quadrilateral add to 180°". A number without a reason often loses marks.
Arc length and sector area (quick link)
A central angle θ (in degrees) cuts off a fraction θ/360 of the circle. So arc length = (θ/360) × 2πr and sector area = (θ/360) × πr². Example: r = 6 cm, θ = 60°: arc = (1/6) × 12π ≈ 6.28 cm, sector = (1/6) × 36π ≈ 18.85 cm². More in the lesson on area of a circle.
Try it at home
Draw a circle round a plate. Draw any diameter. Put the corner of a book on the circle so its edges pass through both ends of the diameter. It always fits exactly: the angle in a semicircle is 90°. Then mark two points A and B, pick three more points on the big arc and measure each angle APB with a protractor. They should all match.
Key formulas and definitions
- Angle at centre = 2 × angle at circumference
- Angle in a semicircle = 90°
- Angles in the same segment are equal
- Cyclic quadrilateral: ∠A + ∠C = 180°, ∠B + ∠D = 180°
- Tangent ⟂ radius at the point of contact
- Tangents from an outside point are equal: ET = √(OE² − r²)
- Alternate segment: tangent–chord angle = angle in the other segment
- Arc length = (θ/360) × 2πr; sector area = (θ/360) × πr²
Worked examples
1. ∠APB at the edge is 35°. Find the angle ∠AOB at the centre on the same chord.
Step 1: angle at centre = 2 × angle at circumference. Step 2: ∠AOB = 2 × 35° = 70°.
2. AB is a diameter and ∠PAB = 52°. Find ∠PBA.
Step 1: ∠APB = 90° (angle in a semicircle). Step 2: angles in triangle add to 180°, so ∠PBA = 180° − 90° − 52° = 38°.
3. ABCD is a cyclic quadrilateral with ∠A = 112° and ∠B = 75°. Find ∠C and ∠D.
Opposite angles add to 180°. ∠C = 180° − 112° = 68°. ∠D = 180° − 75° = 105°.
4. A tangent from E touches a circle of radius 5 cm at T. OE = 13 cm. Find ET.
Step 1: ∠OTE = 90° (tangent ⟂ radius). Step 2: ET² = OE² − OT² = 169 − 25 = 144. Step 3: ET = 12 cm.
5. Points P and Q are on the same arc above chord AB. ∠APB = 48°. ∠AOB is the centre angle. Find ∠AQB and ∠AOB.
∠AQB = 48° (angles in the same segment are equal). ∠AOB = 2 × 48° = 96° (angle at centre is twice the angle at circumference).
6. A tangent at T makes an angle of 64° with chord TA. P is on the major arc. Find ∠TPA and ∠TOA.
Step 1: alternate segment theorem gives ∠TPA = 64°. Step 2: angle at centre is double, so ∠TOA = 128°. Check: triangle OTA is isosceles, its base angles are (180 − 128)/2 = 26°, and 26° + 64° = 90°, which matches tangent ⟂ radius.
7. Two tangents from E touch a circle at A and B. ∠AEB = 50°. Find ∠AOB.
Quadrilateral OAEB has two 90° angles (tangent ⟂ radius). Angles of a quadrilateral add to 360°, so ∠AOB = 360° − 90° − 90° − 50° = 130°.
Common mistakes
- Halving when you should double: the angle at the centre is the BIGGER one, twice the angle at the edge.
- Using the cyclic quadrilateral rule when one corner is the centre O. All four corners must be on the circle.
- Using "same segment" for points on opposite sides of the chord. Points on opposite sides give angles that add to 180°, not equal angles.
- Writing angles without a reason. Name the theorem every time; exams give marks for reasons.