Question

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

Answer 1

Two circle with centre $O$ and $O^{\prime }$ intersect at $A$ and $B$. AB is common chord of two circle $O{O}^{\prime }$ is the line joining centre

Let $O{O}^{\prime }$ intersect $AB$ at $P$

In $OAO$ and $OB{O}^{\prime }$ we have

$O{O}^{\prime }\to$ common

$OA=OB\to$(radii of the same circle)

$O^{\prime }A=O^{\prime }B\to$(radii of the same circle)

$\Rightarrow △OA{O}^{\prime }\cong △OB{O}^{\prime }$ {$SSS$ conguence}

$\angle AO{O}^{\prime }=\angle BO{O}^{\prime }\left(CPCT\right)$

i.e., $\angle AOP=\angle BOP$

In $△AOP$ and $BOP$ we have $OP=OP$ common

$\angle AOP=\angle BOP$ (proved above)

$OA=OB$ (Radii of the semicircle)

$△APD=△BPD$ ($SSS$ conguence)

$AP=CP\left(CPCT\right)$

and $\angle APO=\angle BPO\left(CPCT\right)$

But $\angle APO+\angle BPO=180$

$\therefore \angle APO={90}^o$

$\therefore AP=BP$ and $\angle APO=\angle BPO={90}^o$

$\therefore O{O}^{\prime }$ is perpendicular bisector of $AB$