Question

Prove that the line of centres of two intersecting circles bisects their common chord.

Answer 1

Given : Two circles of radius $r_1\&r_2$ intersect at two different points $A$, $B$ and $PQ$ is $\perp$ bisector of $AB$.

$\therefore$ $AO=OB$ and $\angle B={90}^0$

To prove center of circle $C_1\&C_2$ lie on $Pa$

Construction Join $A$ to $C_1\&C_2$ and also $B$ to $C_1$ and $C_2$

Proof : $C_{1}A=C_{1}B=r$ and also $C_{2}A=C_{2}B=r_2$

$\therefore$ quadrilateral $C_{1}A$ $C_{2}B$ is a hite.

$\because$ $PQ$ is $\perp$ & bisector of $AB$

$\therefore$ $AB$ is shorter diagonal

$\therefore$ $C_1$ and $C_2$ are in $PQ$.