Question

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

Answer 1

Step 1: Prove that $PR=PQ.$

Let circle $C_1$ have Centre $O$ and circle $C_2$ have centre $O\text{'}$,$PQ$ is the common chord.

Construction: Join $PO,PO\text{'},QO,QO\text{'}$

In$△POO\text{'}$ and $△QOO\text{'}$

$OP=OQ$(Radius of circle $C_1$)

$O\text{'}P=O\text{'}Q$(Radius of circle $C_2$)

$OO\text{'}=OO\text{'}$(Common)

$\therefore △POO\text{'}\cong QOO\text{'}$(SSS Congruence rule)

$\angle POO\text{'}=\angle QOO\text{'}$(CPCT) $--\left(1\right)$

Also

In$△POR$ and $△QOR$

$OP=OQ$(Radius of circle $C_1$)

$\angle POR=\angle QOR$(From $\left(1\right)$)

$OR=OR$(Common)

$\therefore △POO\text{'}\cong QOO\text{'}$ (SAS Congruence rule)

$\Rightarrow PR=QR$(CPCT)

& $\angle PRO=\angle QRO$(CPCT) $--\left(2\right)$

Step 2: Prove that $\angle PRO=\angle PRO\text{'}=\angle QRO=\angle QRO\text{'}=90°$

Since $PQ$ is a line.

$\angle PRO+\angle QRO=180°$ (Linear Pair)

$\Rightarrow \angle PRO+\angle PRO=180$ (From $\left(2\right)$)

$\Rightarrow 2\angle PRO=180$

$$\begin{gathered} \Rightarrow \angle PRO=\frac{180}{2} \\ \Rightarrow \angle PRO=90° \end{gathered}$$

Therefore,

$\angle QRO=\angle PRO=90°$

Also,

$\angle PRO\text{'}=\angle QRO=90°$ (Vertically opposite angles)

$\angle QRO\text{'}=\angle PRO=90°$ (Vertically opposite angles)

Since,$\angle PRO=\angle PRO\text{'}=\angle QRO=\angle QRO\text{'}=90°$ and $PR=QR$

we can say that $OO\text{'}$ is the perpendicular bisector of $PQ$

Hence proved that centers of $C_1$ and $C_2$ lie on the perpendicular bisector of the common chord.