Question

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

Answer 1

In $△POX$ and $△QOX$

$OP=OQ$ radius of circle $C_1$

$XP=XQ$ radius of circle $C_2$

$OX=OX$(common)

$\therefore △POX\cong △QOX$ by SSS congruence rule

$\angle POX=\angle QOX$ by C.P.C.T ..$\left(1\right)$

Also, In $△POR$ and $△QOR$

$OP=OQ$ radius of circle $C_1$

$\angle POR=\angle QOR$ since $\angle POX=\angle QOX$

$OR=OR$ (common)

$\therefore △OPX\cong △OQX$ by SAS congruence rule

$PR=QR$ by CPCT

and $\angle PRO=\angle QRO$ by CPCT ....$\left(2\right)$

Since $PQ$ is a line

$\angle PRO+\angle QRO={180}^{\circ }$ (linear pair)

$\angle PRO+\angle PRO={180}^{\circ }$ from $\left(2\right)$

$\Rightarrow 2\angle PRO={180}^{\circ }$

$\Rightarrow \angle PRO=\frac{{180}^{\circ }}{2}={90}^{\circ }$

$\therefore \angle QRO=\angle PRO={90}^{\circ }$

Also, $\angle PRX=\angle QRO={90}^{\circ }$ (vertically opposite angles)

$\angle QRX=\angle PRO={90}^{\circ }$ (vertically opposite angles)

Since $\angle PRO=\angle QRO=\angle PRX=\angle QRX={90}^{\circ }$

$\therefore ,OX$ is the perpendicular bisector of $PQ$