Theorem

The perpendicular from the centre bisects a chord

Geometry • Circle • Chords

Formula

d^{2} + (\frac{chord}{2})^{2} = r^{2}

The perpendicular from the centre of a circle to a chord bisects that chord, and conversely the line from the centre to a chord's midpoint is perpendicular to it; hence for radius

r

, chord length

2 m

and distance

d

from the centre,

d^{2} + m^{2} = r^{2}

How to use it

1Draw the radius to one end of the chord: that radius, half the chord and the distance from the centre form a right triangle.
2A longer chord lies closer to the centre, and the diameter is the unique chord through it.
3Where a diameter meets a chord perpendicularly at $P$ , the two pieces of the diameter are $r + d$ and $r - d$ .

Worked example

A chord of length

16

cm lies in a circle of radius

10

cm. How far is it from the centre?

Step-by-step

1
Half the chord is $8$ cm.
Why: The perpendicular from the centre bisects the chord.
2
$d = 1 0^{2} - 8^{2} = 36 = 6$ cm

Answer

6

Watch out

The right triangle uses half the chord, not the whole of it, and the radius is the hypotenuse — so the distance is

r^{2} - m^{2}

and never

r^{2} + m^{2}

. A distance coming out larger than

r

signals that the subtraction was done the wrong way round.

Builds on

Pythagoras' theorem

Also known as

perpendicular from centre to a chord, chord bisection, distance from centre to a chord, half chord

Appears in past papers

Formula

Worked example

Appears in past papers

2020 (I)

2019 (II)

2019 (I)

2018 (II)

2018 (I)

2017 (II)

2017 (I)

2016 (II)