Question

Prove that the length of the common chord of the two circles whose equations are $(x-a{)}^2+(y-b{)}^2=c^2$ and $(x-b{)}^2+(y-a{)}^2=c^2$ is $\sqrt{4c^2-2(a-b{)}^2}$.
Hence find the condition that the two circles may touch.

Answer 1

As the radius of the given circles are equal so the common chord will bisect the line joining their centres

$O{O}^{\prime }=\sqrt{{(a-b)}^2+{(b-a)}^2}O{O}^{\prime }=\sqrt{{(a-b)}^2+{(a-b)}^2}O{O}^{\prime }=\sqrt{2}|a-b|$

Now $O{O}^{\prime }=2AB$

$\Rightarrow OB=\frac{\sqrt{2}|a-b|}{2}=\frac{|a-b|}{\sqrt{2}}$

In $△AOB$ $,$ ${OA}^2={OB}^2+{AB}^2$

$\Rightarrow c^2={\left(\frac{|a-b|}{\sqrt{2}}\right)}^2+{AB}^2\Rightarrow {AB}^2=c^2-\frac{{(a-b)}^2}{2}\Rightarrow {AB}^2=\frac{2c^2-{(a-b)}^2}{2}\Rightarrow AB=\sqrt{\frac{2c^2-{(a-b)}^2}{2}}=\frac{\sqrt{4c^2-{2(a-b)}^2}}{2}$

Length of common chord $=AC=2AB$

$\Rightarrow AC=2\times \frac{\sqrt{4c^2-{2(a-b)}^2}}{2}=\sqrt{4c^2-{2(a-b)}^2}$

Hence proved.