Question

If two circles intersect at two distinct points, prove that the line through their centres is the perpendicular bisector of their common chord.

Answer 1

Draw two circles with centre $O$ and $P$ which intersects each other at two distinct points $A$ and $B$

Join $AB$ which forms common chord for both the circles and then join $OP$.

In $△OAP$ and $△OBP,$

$OA=OB$ ....... [Radii of same circle]

$AP=BP$ ........ [Radii of same circle]

$OP=OP$ ....... [Common side]

$\therefore △OAP\cong △OBP$ ........ $\left(i\right)$ [By SSS rule]

$⟹\angle AOP=\angle BOP$ ........ [CPCT]

$⟹\angle AOC=\angle BOC$ ........... $\left(ii\right)$

Now, in $△AOC$ and $△BOC$

$OA=OB$ ....... [Radii of same circle]

$\angle AOC=\angle BOC$ ....... From $\left(ii\right)$

$OC=OC$ ........... [Common side]

$⟹△AOC\cong △BOC$

$⟹AC=BC$ .......... [CPCT]

$\angle ACO=\angle BCO$ ........ [CPCT]

But $\angle ACO+\angle BCO={180}^o$

$⟹\angle ACO={90}^o=\angle BCO$

Thus, $OP$ is the perpendicular bisector of $AB$