If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Given:

Two circles intersect at two points

To do:

We have to prove that their centres lie on the perpendicular bisector of the common chord.

Solution:

Let two circles with centres $O$ and $O'$ intersect each other at $A$ and $B$.

$OA = OB$                (Radii of the circle)

$O’A = O'B$               (Radii of the circle)

$OO’ = OO’$              (Common side)

Therefore, by SSS congruency,

$\triangle AOO’$ and $\triangle BOO’$ are similar.

This implies,

$\triangle AOO’ \cong \triangle BOO’$

$\angle AOO’ = \angle BOO’$............(i)

In $\triangle AOC$ and \triangle BOC$,

$OA = OB$         (Radii)

$\angle AOC = \angle BOC$          ($\angle AOO’ = \angle BOO’$)

$OC = OC$             (Common side)

Therefore, by SAS congruency,

$\triangle AOC \cong \triangle BOC$

This implies,

$\angle ACO = \angle BCO$

$\angle ACO+\angle BCO = 180^o$

$2\angle ACO=180^o$

$\angle ACO = \angle BCO = \frac{180^o}{2} = 90^o$

$OO’$ is the perpendicular bisector of $AB$.

Hence, their centres lie on the perpendicular bisector of the common chord.

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Updated on: 10-Oct-2022

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