The length of common chord of the circles $x^2+y^2-6 x-4 y+13-c^2=0$ and $x^2+y^2-4 x-6 y+13-c^2=0$ is

• Correct answer
  A

  $\sqrt{4 c^2-2}$
• B

  $\frac{1}{2} \sqrt{4 c^2-2}$
• C

  $\sqrt{c^2-2}$
• D

  $\sqrt{4 c^2-1}$

Solution

Given circles, $$ \begin{aligned} & x^2+y^2-6 x-4 y+13-c^2=0 \text { and } \\ & x^2+y^2-4 x-6 y+13-c^2=0 \end{aligned} $$ Centres of the circles are $C_1(3,2)$ and $C_2(2,3)$ and $r_1=c$ and $r_2=c$ The circles intersects at two points. Radical axis is the common chord of the circle equation of common chord is $G_1-G_2=0$ $$ \therefore-2 x+2 y=0 \Rightarrow x-y=0 $$ Draw $C_1 M \perp A B$
$\begin{aligned} & C_1 M=\left|\frac{3-2}{\sqrt{(1)^2+(-1)^2}}\right| \Rightarrow C_1 M=\frac{1}{\sqrt{2}} \\ & \text { In } \triangle A C_1 M \\ & A M^2=\left(A C_1\right)^2-\left(C_1 M\right)^2=C^2-\left(\frac{1}{\sqrt{2}}\right)^2 \\ & A M^2=c^2-\frac{1}{2} \Rightarrow A M=\sqrt{c^2-\frac{1}{2}} \\ & \text { So, } A B=2 A M=2 \sqrt{c^2-\frac{1}{2}} \\ & \Rightarrow \sqrt{4 c^2-2}\end{aligned}$

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