Cambridge IGCSE Mathematics 0580: Circle theorems II – Study Guide

This study guide teaches subtopic E4.8, Circle theorems II, from the Cambridge IGCSE Mathematics 0580 syllabus for 2025, 2026 and 2027. The whole subtopic is Extended only: the syllabus lists C4.8 as “Extended content only”, so there is no Core version. It is examined on Paper 2 (non-calculator) and Paper 4 (calculator), the two Extended papers.

Use it with the 0580 course hub and the printable 0580 checklist. The angle theorems of C4.7 and E4.7 are covered in the geometry study guide, with geometry practice questions to test them. To check the whole Extended course, try the Extended diagnostic.

What this subtopic covers

The syllabus asks you to use three symmetry properties of circles, and to quote them as reasons for your answers.

0580 reference	What you must be able to do	Tier
E4.8	Use the fact that equal chords are equidistant from the centre	Extended only
E4.8	Use the fact that the perpendicular bisector of a chord passes through the centre	Extended only
E4.8	Use the fact that tangents from an external point are equal in length	Extended only
E4.8 (notes)	Give reasons using the geometrical properties listed in the syllabus	Extended only

Most E4.8 questions also use Pythagoras’ theorem, right-angled trigonometry, isosceles triangles and the fact that a tangent meets a radius at 90° (C4.7).

The symmetry behind the properties

Join the centre O to the ends of a chord AB. OA and OB are radii, so triangle OAB is isosceles. Its line of symmetry passes through O and meets AB at its midpoint at 90°. That line is the perpendicular bisector of AB, so it contains O, and the perpendicular from O to a chord cuts the chord in half. You do not need to prove this, but it shows you which right-angled triangle to work in. The distance from the centre to a chord always means this perpendicular distance. For tangents from P, triangles OAP and OBP are right-angled at A and B, share the hypotenuse OP and have OA = OB, so Pythagoras gives PA = PB.

Property 1: the perpendicular bisector of a chord passes through the centre

In practice you use this property in two ways.

1. Lengths. Draw OM perpendicular to chord AB. Then M is the midpoint of AB, and triangle OMA is right-angled at M with hypotenuse OA = r. Pythagoras gives r² = OM² + (AB/2)².
2. Finding the centre. Any two chords that are not parallel have perpendicular bisectors that cross at exactly one point. Both bisectors pass through the centre, so that point is the centre.

Worked example 1 (non-calculator)

A circle has centre O and radius 8.5 cm. AB is a chord of length 15 cm. Find the distance from O to AB.

Draw OM perpendicular to AB.
The perpendicular from the centre bisects the chord, so AM = 15 ÷ 2 = 7.5 cm.
Triangle OMA is right-angled at M:
  OM² = OA² − AM² = 8.5² − 7.5²
      = (8.5 − 7.5)(8.5 + 7.5) = 1 × 16 = 16
  OM = 4 cm

The difference of two squares keeps the arithmetic easy without a calculator.

Worked example 2 (calculator)

A horizontal cylindrical tank is partly filled with water. In the circular cross-section, centre O, the water surface AB is 24 cm wide and the water is 8 cm deep at its deepest point. The tank is less than half full.

(a) Find the radius of the tank. (b) Find angle AOB, the angle the water surface subtends at the centre.

(a) Let the radius be r. The deepest point lies on the line through O
    perpendicular to AB (the perpendicular bisector of a chord passes
    through the centre). Call the midpoint of AB M.
    AM = 24 ÷ 2 = 12 cm
    OM = r − 8   (the tank is less than half full, so O is above the water)
    In triangle OMA:  r² = (r − 8)² + 12²
                      r² = r² − 16r + 64 + 144
                      16r = 208
                      r = 13 cm

(b) sin(angle AOM) = AM / OA = 12/13
    angle AOM = 67.38...°
    angle AOB = 2 × 67.38...° = 134.8°  (1 d.p.)

Keep 67.38…° in your calculator and double it before rounding. Early rounding happens to give the same answer here, but it can cost accuracy in other questions.

Worked example 3 (non-calculator): finding the centre with coordinates

A circle passes through A(1, 1), B(7, 1) and C(5, 5). Find the coordinates of its centre and its radius.

Perpendicular bisector of AB:
  AB is horizontal, midpoint (4, 1), so the bisector is x = 4.

Perpendicular bisector of BC:
  midpoint of BC = ((7 + 5)/2, (1 + 5)/2) = (6, 3)
  gradient of BC = (5 − 1)/(5 − 7) = −2
  perpendicular gradient = 1/2
  y − 3 = (1/2)(x − 6),  so  y = (1/2)x

Both bisectors pass through the centre:
  x = 4  gives  y = 2.   Centre (4, 2).

Radius = distance from (4, 2) to A(1, 1) = √(3² + 1²) = √10
Check with C(5, 5): √(1² + 3²) = √10

This uses the perpendicular-bisector skills from coordinate geometry. Checking the third point catches slips.

Property 2: equal chords are equidistant from the centre

If two chords of the same circle have the same length, their perpendicular distances from the centre are equal. The reason is Pythagoras again: the same radius and the same half-chord give the same third side. The reverse also follows from the same calculation: chords the same distance from the centre have the same length.

Worked example 4 (non-calculator)

AB and CD are parallel chords of a circle with centre O and radius 6.5 cm. AB = CD = 12 cm, and the chords are not the same line.

(a) Find the distance between the two chords. (b) Find the area of quadrilateral ABDC.

(a) Distance from O to AB:
      √(6.5² − 6²) = √(42.25 − 36) = √6.25 = 2.5 cm
    Equal chords are equidistant from the centre, so O is also 2.5 cm from CD.
    Two different parallel chords 2.5 cm from O must lie on opposite sides of O.
    Distance between the chords = 2.5 + 2.5 = 5 cm

(b) AB and CD are equal and parallel, and the line joining their midpoints
    is perpendicular to both, so ABDC is a rectangle.
    Area = 12 × 5 = 60 cm²

A quick check: the diagonal of the rectangle is √(12² + 5²) = 13 cm, which is the diameter. That fits the angle in a semicircle: each corner of the rectangle is 90°, so each diagonal must be a diameter.

The key step is in (a): the question does not say the chords are on opposite sides of O, so you have to work it out.

Property 3: tangents from an external point are equal in length

From an external point P, draw tangents touching the circle at A and B. Then PA = PB. Two consequences come up again and again.

• Triangle PAB is isosceles, so angle PAB = angle PBA.
• Quadrilateral OAPB is a kite (OA = OB as radii, PA = PB as tangents), with right angles at A and B because the tangent meets the radius at 90°. OP is its line of symmetry.

Worked example 5 (non-calculator, giving reasons)

PA and PB are tangents from P to a circle with centre O. Angle APB = 48°. C is a point on the major arc AB.

Find, giving a reason for each step: (a) angle PAB, (b) angle AOB, (c) angle ACB.

(a) PA = PB  (tangents from an external point are equal in length)
    so triangle PAB is isosceles.
    angle PAB = (180° − 48°) ÷ 2 = 66°
    (base angles of an isosceles triangle; angle sum of a triangle = 180°)

(b) angle OAP = angle OBP = 90°  (angle between tangent and radius = 90°)
    angle AOB = 360° − 90° − 90° − 48° = 132°
    (angle sum of a quadrilateral = 360°)

(c) angle ACB = 132° ÷ 2 = 66°
    (angle at the centre is twice the angle at the circumference)

Notice that angle ACB = angle PAB. That is the alternate segment theorem from E4.7, so it is a useful check on your arithmetic.

Worked example 6 (calculator)

From a point P, 17 cm from the centre O of a circle of radius 8 cm, tangents PA and PB are drawn.

(a) Find the length of PA. (b) Find the area of kite OAPB. (c) Find angle APB.

(a) angle OAP = 90° (tangent meets radius at 90°)
    PA = √(17² − 8²) = √225 = 15 cm

(b) PB = PA = 15 cm (tangents from an external point are equal)
    Area = 2 × (1/2 × 8 × 15) = 120 cm²

(c) tan(angle APO) = OA / PA = 8/15
    angle APO = 28.07...°
    OP is the line of symmetry of the kite, so
    angle APB = 2 × 28.07...° = 56.1°  (1 d.p.)

Worked example 7 (non-calculator): a circle inside a triangle

A circle touches all three sides of triangle ABC. AB = 10 cm, BC = 12 cm and CA = 8 cm. Find the length of the tangent from each vertex to the circle.

Let the tangent lengths from A, B and C be a, b and c.
Each vertex is an external point, so its two tangents are equal:
  AB:  a + b = 10
  BC:  b + c = 12
  CA:  c + a = 8
Add all three:  2(a + b + c) = 30,  so  a + b + c = 15
  a = 15 − 12 = 3 cm
  b = 15 − 8  = 7 cm
  c = 15 − 10 = 5 cm

Check: 3 + 7 = 10, 7 + 5 = 12 and 5 + 3 = 8. Labelling each tangent length once and using it on both sides is the whole method.

Giving reasons

The syllabus notes say you are expected to use the geometrical properties listed in the syllabus when giving reasons. Write them in full. These wordings match the syllabus list:

• “the perpendicular bisector of a chord passes through the centre”
• “equal chords are equidistant from the centre”
• “tangents from an external point are equal in length”
• “angle between tangent and radius = 90°”

Short forms such as “tangent rule” or “chord theorem” do not say which property you used. When a question says “give a reason”, a correct number with no reason, or with a vague one, may not earn full credit.

Paper 2 and Paper 4

• Paper 2 (non-calculator): you work everything by hand. Use tricks such as the difference of two squares, give exact surd answers such as 2√5 when asked for an exact value (surds are E1.18, so simplify them), and set out angle chains like worked example 5 with a reason on every line.
• Paper 4 (calculator): you can use trigonometry inside the right-angled triangles, as in worked examples 2 and 6. Keep full calculator values until the end, then round to 3 significant figures, or 1 decimal place for angles.

Common errors

• Using the whole chord as a side of the right-angled triangle. The triangle uses half the chord.
• Measuring the distance from the centre to the chord along a slanted line. It must be the perpendicular distance.
• Taking the tangent length as OP. The tangent length is PA, from the external point to the point of contact.
• Assuming two equal chords are parallel or on opposite sides of O when the question has not said so. Work it out from what you are given.
• Forgetting the right angle between tangent and radius before using Pythagoras in triangle OAP.
• Using OM = r − depth when the tank is more than half full. Then the centre is under the water surface and OM = depth − r. Draw the diagram first.
• Doubling an angle that has already been rounded, then rounding again.

Next steps

Use the geometry revision notes for a list of every circle theorem for your tier. Then try the chord question in the Extended trigonometry practice questions, which uses the half-chord triangle from worked examples 1 and 2. The course hub lists other resources for syllabus 0580.

Official syllabus

Cambridge International, Cambridge IGCSE Mathematics 0580 syllabus for 2025, 2026 and 2027, Subject content, section 4 Geometry, E4.8 Circle theorems II (Extended subject content), with C4.8 marked “Extended content only”.