The circle $S=0$ cuts the circles $C_1=x^2+y^2-8 x-2 y+16=0$ and $C_2=x^2+y^2-4 x-4 y-1=0$ orthogonally. If the common chord of $S=0$ and $C_1=0$ is $2 x+13 y-15=0$, then the centre of $S=0$ is

• Correct answer
  A

  $\left(\frac{-11}{3}, \frac{7}{6}\right)$
• B

  $\left(\frac{11}{3}, \frac{-7}{6}\right)$
• C

  $\left(\frac{2}{13}, \frac{11}{15}\right)$
• D

  $\left(\frac{11}{15}, \frac{-2}{13}\right)$

Solution

Let the equation of circle $S$ is $\begin{aligned} & S=x^2+y^2+2 g x+2 f y+c=0 \\ & C_1=x^2+y^2-8 x-2 y+16=0 \\ & C_2=x^2+y^2-4 x-4 y-1=0 \end{aligned}$ $S$ is orthognoally to $C_1$ and $C_2$
From Eqs. (i) and (ii),
Common chord of $S$ and $C_1=$
Given common chord is
Eqs. (iv) and (v) are coinside $\therefore \quad \frac{2 g+8}{2}=\frac{2 f+2}{13}=\frac{c-16}{-15}$
Solving Eqs. (iii) and (vi), we get $g=\frac{11}{3} f=\frac{-7}{6}$ $\therefore$ Centre of circle, $S=\left(\frac{-11}{3}, \frac{7}{6}\right)$

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