Chords - Higher

The perpendicular from the centre of a circle to a chord bisects the chord.

Circle with chord bisected with line from the centre

Example

A circle has a radius of 5 cm. The chord EF is 7 cm.

How far is the midpoint of the chord from the centre of the circle?

A circle with the perpendicular from the centre that bisects the chord positioned between 12pm and 2:30pm with the labels F and E.

Add the radii, OE and OF, to make two right-angled triangles.

Circle with radius of 5 cm and a chord labelled FE as 7 cm with mid-point of 3.5 cm. Two radii OE and OF form 2 right-angled triangles. One right-angled triangle is shown as OF 5 cm and FM 3.5 cm.

FM is half of the length of chord EF.

FM = 3.5 cm

Use Pythagoras' theorem to calculate the length OM.

OF2 = FM2 + OM2

OM2 = 5^2 - 3.5^2

OM2 = 12.75

OM = 3.6 cm (to 1 decimal place)

Proof

In the diagram below, AB is the chord of a circle with centre O.

OM is the perpendicular from the centre to the chord.

In the diagram, AB is the chord of a circle with centre O. OM is the perpendicular from the centre to the chord.

Angles OMA and OMB are both right-angles.

OA is the hypotenuse of triangle OAM.

OB is the hypotenuse of triangle OBM.

OA = OB as both are radii of the circle.

OM is common to both triangles.

Therefore, triangles OAM and OMB are congruent (RHS – right-angle, hypotenuse, side).

Therefore, the remaining sides of the triangles are equal, AM = MB.

So, M must be the mid-point of AB, and the chord has been bisected.