The length of the common chord of the two circles $(x-a)^2+(y-b)^2=c^2$ and $(x-b)^2+(y-a)^2=c^2$ is

• A

  $\sqrt{4 c^2+2(a-b)^2}$
• B

  $\sqrt{4 c^2-(a-b)^2}$
• Correct answer
  C

  $\sqrt{4 c^2-2(a-b)^2}$
• D

  $\sqrt{2 c^2-2(a-b)^2}$

Solution

The equations of two circles are $S_1 \equiv(x-a)^2+(y-b)^2=c^2...(i)$ and $\quad S_2 \equiv(x-b)^2+(y-a)^2=c^2...(ii)$ The equation of the common chord of these circle is
$\begin{aligned} & \Rightarrow(x-a)^2-(x-b)^2+(y-b)^2-(y-a)^2=0 \\ & \Rightarrow(2 x-a-b)(b-a)+(2 y-b-a)(a-b)=0 \\ & \Rightarrow 2 x-a-b-2 y+b+a=0 \\ & \Rightarrow x-y=0\end{aligned}$ Let $C_1$ and $C_2$ be the centres of circles (i) and (ii) respectively. Then, the coordinates of $C_1$ and $C_2$ are $(a, b)$ and $(b, a)$ respectively. $C_1 M=\left|\frac{a-b}{\sqrt{1+1}}\right|=\frac{|a-b|}{\sqrt{2}}$ In right $\Delta C_1 P M$, we have, $\begin{aligned} P M & =\sqrt{C_1 P^2-C_1 M^2}=\sqrt{c^2-\frac{(a-b)^2}{2}} \\ \therefore P Q & =2 P M =2 \sqrt{c^2-\frac{(a-b)^2}{2}} \\ & =\sqrt{4 c^2-2(a-b)^2} \end{aligned}$

Hence, the correct answer is option: 'C'.