Lesson 6: Circles and Theorems

Circles are everywhere in British life — the M25, Big Ben's clock face, the London Eye, roundabouts from Swindon to Milton Keynes. GCSE circle work has two halves: measurement (circumference, area, arcs, sectors) and the circle theorems — the elegant facts about angles in circles that Higher-tier papers test with proof-style questions. Master both: the formulas for accuracy, the theorems with their exact wordings for marks.

Circumference and area

For a circle of radius r: circumference C = 2πr (or πd using the diameter), area A = πr². For a wheel of radius 30 cm: C = 2π(30) = 60π ≈ 188.5 cm — the distance per revolution; A = π(30²) = 900π ≈ 2,827 cm². Learn the inverse problems too: given C = 100 cm, the radius is 100 ÷ 2π ≈ 15.9 cm. GCSE often wants answers 'in terms of π' (exact form, e.g. 60π cm) — one mark for the method, one for leaving π symbolic.

Worked example: the London Eye has a diameter of 120 m. How far does a capsule travel in one full revolution? C = πd = π × 120 = 120π ≈ 377 m. And the area enclosed by the wheel's rim: A = π × 60² = 3600π ≈ 11,310 m². (Watch which length you are given: the Eye's 'diameter 120 m' means r = 60 m — the most common slip in this topic is using the diameter in the area formula.)

Real context: a garden in Kent needs turf for a circular lawn of radius 5 m. Turf sells in 1 m² rolls at £4.50 each. Area = π × 25 = 25π ≈ 78.5 m², so you need 79 rolls (round up — you cannot buy part of a roll in this model, and 78 rolls leave a gap), costing 79 × 4.50 = £355.50. GCSE context questions reward the sensible rounding decision, not just the multiplication.

Arcs and sectors

A sector is a pizza slice: a region bounded by two radii and an arc. For an angle θ at the centre: arc length = (θ/360°) × 2πr; sector area = (θ/360°) × πr². The fraction θ/360 is the whole method — the sector is that fraction of the full circle.

Worked example: a sector has radius 12 cm and angle 75°. Arc length = (75/360) × 2π(12) = (5/24) × 24π = 5π ≈ 15.7 cm. Sector area = (75/360) × π(12²) = (5/24) × 144π = 30π ≈ 94.2 cm².

Worked example: a semicircular window has radius 0.8 m. Its perimeter (arc + diameter) = π(0.8) + 1.6 = 0.8π + 1.6 ≈ 2.51 + 1.6 = 4.11 m. Forgetting the straight edge in a perimeter question is the classic sector error — area questions never need it, perimeter questions usually do.

Segments (the region between a chord and an arc) need the triangle subtracted: segment area = sector area − triangle area. For a segment with angle 90° and radius 10 cm: sector = (90/360) × π × 100 = 25π ≈ 78.5 cm²; the triangle (two radii and the chord) is right-angled with legs 10, 10, so area = ½ × 10 × 10 = 50 cm²; segment = 78.5 − 50 = 28.5 cm².

The circle theorems

There are eight to know. Learn each with its exact wording — GCSE mark schemes quote them: 1. The angle at the centre is twice the angle at the circumference. (Both standing on the same chord.) 2. Angles in the same segment are equal. 3. The angle in a semicircle is a right angle. 4. Opposite angles in a cyclic quadrilateral sum to 180°. 5. The radius to a tangent is perpendicular (a tangent touches the circle at exactly one point; the radius to that point meets it at 90°). 6. Tangents from an external point are equal in length. 7. The perpendicular from the centre to a chord bisects the chord. 8. Alternate segment theorem: the angle between a tangent and a chord equals the angle in the alternate segment.

Worked example: O is the centre; angle AOB = 116° at the centre, with C on the circumference on the major arc. Angle ACB = ½ × 116° = 58° (theorem 1). If instead you need the reflex angle at O for a point on the minor arc: reflex AOB = 360° − 116° = 244°, giving an angle at the circumference of 122°.

Worked example: AB is a diameter; C is on the circle with angle BAC = 37°. Angle BCA = 90° (angle in a semicircle, theorem 3), so angle ABC = 180° − 90° − 37° = 53° (angle sum of a triangle). Two theorems, chained — the standard GCSE pattern.

Worked example: ABCD is cyclic with angle BAD = 105°. Angle BCD = 180° − 105° = 75° (theorem 4). And with a tangent PT touching at T, where chord TA makes angle PTA = 42° with the tangent, the angle in the alternate segment (the angle in the opposite arc on TA) is also 42° (theorem 8).

Problem solving with circle theorems

The method is the same as angle-chasing everywhere: mark what you know, then hunt theorem by theorem. Useful habits: draw in the radius to any tangent point (instant right angle); join the centre to points on the circumference to make isosceles triangles (two radii are equal — base angles equal); look for diameters (instant right angles at the circumference).

Worked example: from an external point P, two tangents touch a circle at A and B; the angle at the centre AOB is 130°. Find angle APB. The quadrilateral OAPB has right angles at A and B (theorem 5, twice), so angle APB = 360° − 90° − 90° − 130° = 50°. Notice the pattern: angle APB = 180° − angle AOB, always — tangents and the centre form a kite-like quadrilateral.

Worked example: a chord of length 16 cm lies 6 cm from the centre. Find the radius. The perpendicular from the centre bisects the chord (theorem 7), giving a right-angled triangle with legs 8 and 6 and hypotenuse r: r² = 8² + 6² = 64 + 36 = 100, so r = 10 cm. Circle theorems and Pythagoras are frequent partners.

Proof expectations (Higher): you may be asked to prove theorem 1 or 3. The standard proof of 'angle in a semicircle': join the centre O to C; OA = OB = OC (radii), making two isosceles triangles; let their base angles be a and b; the angle sum gives 2a + 2b = 180°, so the angle at C, a + b, equals 90°. Write proofs as clean chains of reasons.

Problem solving — arcs, sectors, and a running track

Worked example: a school running track has two straights of 80 m and two semicircular ends of radius 18 m. The full lap: the two semicircles make one complete circle, circumference 2π × 18 = 36π ≈ 113.1 m. Total lap = 80 + 80 + 113.1 = 273.1 m. Runners in the outside lane (radius 19 m) run 2π × 19 ≈ 119.4 m of curve — 6.3 m more, which is exactly why athletics uses staggered starts. Recognising 'two semicircles = one circle' is the simplification the question is built around.

Worked example: a circular pond of radius 4 m is surrounded by a path 1 m wide. The path is an annulus (ring): outer radius 5 m, inner radius 4 m. Path area = π × 5² − π × 4² = 25π − 16π = 9π ≈ 28.3 m². If paving costs £38 per m², the path costs 28.3 × 38 ≈ £1,074. Ring-area problems appear constantly (washers, running tracks, tree surrounds): always outer circle minus inner circle, never π(R − r)² — that tempting shortcut is wrong.

Worked example: a sector has area 50 cm² and radius 10 cm. Find its angle. Sector area = (θ/360) × π × 100 = 50 → (θ/360) = 50/(100π) = 1/(2π) → θ = 360/(2π) = 180/π ≈ 57.3°. Reverse problems — given the area, find the angle or radius — test the same fraction θ/360 working backwards.

Common mistakes

• Diameter in the area formula. A = πr² needs the radius: halve the diameter first.
• Perimeter of a sector without the radii. Arc + 2r for a full sector; arc + diameter for a semicircle.
• Vague theorem wording. 'Angle at centre is double' is not enough — write 'the angle at the centre is twice the angle at the circumference'.
• Missing the reflex angle. If the point sits on the minor arc, the centre angle to use is the reflex one (360° − θ).
• Assuming a line is a diameter or a tangent. Theorems 3 and 5 need the line to *be* a diameter/tangent — check the diagram labels and the question text.
• Cyclic quadrilateral confusion. All four vertices must lie on the circumference for theorem 4 to apply.

Exam Experience

Question 1. A sector of a circle has radius 6 cm and angle 120°. Work out the length of the arc of the sector, giving your answer in terms of π. [2 marks]

Question 2. O is the centre of a circle. A, B and C are points on the circumference. Angle AOC = 146°. Work out the size of angle ABC, giving a reason for your answer. [2 marks]

Question 3. AB is a diameter of a circle. C is a point on the circle. Angle BAC = 37°. Show that angle ABC = 53°. [3 marks]

Mark scheme.

• Q1: M1 for (120/360) × 2π × 6. A1 for 4π cm (in terms of π, as asked).
• Q2: M1 for halving 146°. A1 for 73° with the reason 'the angle at the centre is twice the angle at the circumference'. (The reason is required for full marks.)
• Q3: M1 for stating angle ACB = 90°, with the reason 'angle in a semicircle'. M1 for using the triangle angle sum: angle ABC = 180° − 90° − 37°. A1 for 53°, completing the 'show that'.