Chord Properties

Use chord facts to find lengths: chords crossing inside a circle, two lines from an outside point, and the perpendicular from the centre that cuts a chord in half.

In the exam for Edexcel IGCSE Maths 4MA1, Cambridge IGCSE Maths 0580 · 9 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

Three Chord Facts

A chord joins two points on a circle. When chords cross, or meet outside, their lengths are linked.

Where	Rule
Cross inside at P	part × part, same for both chords
Meet outside at P	outside part × whole, same for both lines
Line from O at 90°	cuts the chord in half

Key idea

Every rule measures from the point P where the lines meet.

Chords That Cross Inside

The 2 or 3-mark question: chords AB and CD cross at P. Multiply the two parts of each chord.

Formula

A P \times P B = C P \times P D

Worked example

1
$A P = 8, P B = 3, C P = 6$ — The lengths you are given, in cm. PD is the one you want.
2
$A P \times P B = C P \times P D$ — The chords cross inside the circle, so part times part of one chord equals part times part of the other.
3
$8 \times 3 = 6 \times P D$ — Put the numbers in. Each side holds the two parts of one chord.
4
$24 = 6 \times P D$ — Work out the left side, 8 × 3.
5
$P D = 24 \div 6$ — PD is multiplied by 6, so divide both sides by 6.
6
$P D = 4 cm$ — Check: 6 × 4 = 24, the same as 8 × 3.

Common mistake

Multiplying the parts of the same chord on both sides. Each side of the equation belongs to one chord.

Two Lines From Outside

The 5 or 6-mark question: lines from P outside the circle cut it at A and B, and at C and D. Use the whole length from P.

Formula

P A \times P B = P C \times P D

Worked example

1
$P A = 4, A B = 5, P C = 3$ — The lengths you are given, in cm. You want the chord CD.
2
$P B = 4 + 5 = 9$ — The rule needs the whole line from P to B. Add the outside part PA and the chord AB.
3
$P A \times P B = P C \times P D$ — The lines meet outside the circle, so outside part times whole line is the same for both lines.
4
$4 \times 9 = 3 \times P D$ — Put the numbers in. On the left, the outside part is 4 and the whole line is 9.
5
$36 = 3 \times P D$ — Work out 4 × 9.
6
$P D = 36 \div 3 = 12$ — Divide both sides by 3. This is the whole line from P to D.
7
$C D = 12 - 3 = 9 cm$ — The chord CD is only the part inside the circle, so take off the outside part PC = 3.

Common mistake

Using the chord $A B = 5$ instead of the whole $P B = 9$ . Outside, always multiply by the whole line from P.

Chord Lengths With Algebra

Exams often put x in the lengths. Set up part × part, multiply out, and solve the quadratic.

Worked example

1
$A P = x, P B = x + 2, C P = 3, P D = 8$ — Lengths in cm. The chords AB and CD cross at P, and x is in the first chord.
2
$x (x + 2) = 3 \times 8$ — Part times part for each chord. The brackets keep the whole length x + 2 together.
3
$x^{2} + 2 x = 24$ — Multiply out the bracket on the left, and work out 3 × 8 on the right.
4
$x^{2} + 2 x - 24 = 0$ — Take 24 from both sides so one side is 0. This is a quadratic equation.
5
$(x + 6) (x - 4) = 0$ — Factorise: the two numbers that multiply to −24 and add to +2 are +6 and −4.
6
$x = - 6 or x = 4$ — Each bracket could be the one that equals 0.
7
$x = 4 cm$ — A length cannot be negative, so reject −6.
8
$4 \times 6 = 24 = 3 \times 8$ — Check: AP × PB with x = 4 is 4 × 6, and CP × PD is 3 × 8. Both are 24.

Common mistake

Writing $x \times x + 2$ and forgetting the brackets. The whole length is $x + 2$ , so it must be $x (x + 2)$ .

Key idea

Write the equation before you simplify it. A correct equation earns marks for working even if the algebra goes wrong later.

Radius From a Diameter

A diameter is a chord that passes through the centre. When it is one of the crossing chords, find its two parts, add them, then halve.

Worked example

1
$C D is a diameter; A P = 6, P B = 3, C P = 2$ — Lengths in cm. You want the radius, which is half of CD.
2
$6 \times 3 = 2 \times P D$ — The chords cross at P, so part times part for each chord.
3
$18 = 2 \times P D$ — Work out 6 × 3.
4
$P D = 18 \div 2 = 9$ — Divide both sides by 2.
5
$C D = 2 + 9 = 11$ — The diameter is the two parts CP and PD end to end.
6
$r = 11 \div 2 = 5.5 cm$ — The radius is half the diameter.

Common mistake

Giving 11 as the radius. 11 is the whole diameter, so halve it as the last step.

The Perpendicular From the Centre

A line from the centre O that meets a chord at $9 0^{\circ}$ cuts the chord into two equal halves.

Rule

Reason: the perpendicular from the centre to a chord bisects the chord.

Worked example

1
$r = 10, distance = 6$ — The radius and the distance from O to the chord, in cm. That distance is measured at 90° to the chord.
2
$half^{2} + 6^{2} = 10^{2}$ — Half the chord, the 6 and the radius make a right-angled triangle. The radius is the longest side, the hypotenuse.
3
$half^{2} = 100 - 36 = 64$ — Take 6² away from both sides to find half squared.
4
$half = 8$ — Square root of 64.
5
$chord = 2 \times 8 = 16 cm$ — The perpendicular from the centre bisects the chord, so the whole chord is double the half.

Common mistake

Stopping at 8. That is only half the chord.

Rule

Chords of equal length are the same distance from the centre, and chords the same distance from the centre are equal. Reason: equal chords are equidistant from the centre.

The same triangle also works backwards. Here the chord is known and the distance is not.

Worked example

1
$chord = 24, r = 13$ — A 24 cm chord in a circle of radius 13 cm. You want how far the chord is from O.
2
$half = 24 \div 2 = 12$ — The perpendicular from O cuts the chord in half, so halve it first.
3
$d^{2} + 1 2^{2} = 13^{2}$ — The distance d, the half-chord and the radius make a right-angled triangle. The radius is the hypotenuse.
4
$d^{2} = 169 - 144 = 25$ — Take 12² away from both sides.
5
$d = 5 cm$ — Square root. A distance cannot be negative, so only 5.

Key idea

Show every line. Most of the marks are for the working, so an answer with no equation behind it can score very little.

Which Method? Read the Question

Find the point P where the lines meet. Is it inside or outside the circle?

You see	You do
Chords cross inside	part × part, same for both
Lines meet outside	outside part × whole, both lines
Unknown in both lengths	expand, then solve the quadratic
Line from O at 90° to a chord	halves, then Pythagoras
Diameter crosses a chord, find r	find both parts, add for the diameter, then halve
Chord and radius, find the distance from O	halve the chord, then Pythagoras with r as hypotenuse
Angles, not lengths	Combining Circle Theorems

Key idea

Draw the lengths on the diagram first. Most lost marks come from using the wrong piece of a line.

Practise chord properties now

Interactive visual aids, then real past-paper questions marked on your working like the exam. Free to start.

Practise free on Lemmara

Real past-paper questions on chord properties

Each with a full worked solution.

All 9 questions on this topic →

Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education or Pearson Edexcel. The lessons and worked examples are Lemmara’s own.