Question

Prove that the line joining the centres of two intersecting circles is the perpendicular bisector of the line joining the points of intersection.

Answer 1

Let A and B be centres of two circles intersecting at points P and Q.

In $△APB$ and $△AQB,$

$AP=AQ\because$They are radii of a circle.

$BP=BQ\because$They are radii of a circle.

$AB=AB\because$Common side

$△APB\cong △AQB$ By S.S.S. Criterion

$\Rightarrow \angle PAB=\angle QAB$ By C.P.C.T.C.

Now, In $△APR$ and $△AQR$

$AP=AQ\because$ They are radii of a circle.

$\angle PAR=\angle QAR\because \angle PAB=\angle QAB$

$AR=AR\because$ Common side

$△APR\cong △AQR$ ....... By S.A.S. Criterion

$\Rightarrow PR=RQ$ ..... By CPCT

And $\angle ARP=\angle ARQ$ ....... By C.P.C.T.C.

Also, $\angle ARP+\angle ARQ={180}^o\because$ PQ is a straight line.

$\Rightarrow \angle ARP+\angle ARP={180}^o$

$\Rightarrow 2\times \angle ARP={180}^o$

$\Rightarrow \angle ARP=\frac{{180}^o}{2}={90}^o$

Thus, $AB{\perp }^{r}PQ$