Chords of circle:

1. Perpendicular bisector of chord
The Perpendicular bisector of chord of a circle passes through the centre of the circle.

A circle is symmetrical about every diameter. Hence any chord AB perpendicular to a diameter is bisected by the diameter.
Also, any chord bisected by a diameter is perpendicular to the diameter.














This property enables us to locate the centre of a circle. The diagram bellows shows a tracing of a broken circular plate. We locate the centre of the plate by marking three points A, B and C on it and drawing the perpendicular bisector of AB and BC. The intersecting point of the perpendicular point of the perpendicular bisectors is the centre.














Proof:

Given a circle, centre O and a chord, SR, with a mid-point Z, we are required to show that OZR = 90°.

ZJoin OS and OR. In triangle OSZ and ORZ,
OS = OR (radii of circle)
SZ = RZ (given)
OZ is common.

Triangle OZS is congruent to triangle OZR (SSS property)

OZR = OZR

Since these are adjacent angles on a straight line, OZR = OZR = 90°

Example:
In the diagram, M is the midpoint of the chord AB and PQ is the perpendicular bisector of AB. AB=8 cm and MQ=2cm. Find the radius of the circle.













Let r be the radius of the circle.

The centre O lies on the line PQ. (Perpendicular bisector of chord)

Join OB as shown in the diagram.

OM square= OM square + MB square

r square =(r-2) square + 4 square
r square = r square – 4r + 4 + 16
4r = 20
r = 5
The radius of the circle is 5 cm.