Question

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 $S=0$

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

Answer 1

Answer: (B)

Let the equation of circle $S$ is
$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 $
$C_{2}=x^{2}+y^{2}-4 x-4 y-1=0$
matheal $S$ is orthogonally to $C_{1}$ and $C_{2}$
$\therefore -8 g-2 f=C+16 \ldots$ (i)
$-4 g-4 f=C-1 \ldots . .$ (ii)
From Eqs. (i) and (ii),
$-4 g+2 f=17 \ldots$ (iii)
Common chord of $S$ and $C_{1}$
=$(2 g+8) x+(2 f+2) y+c-16=0 \ldots$.(iv)
Given common chord is
$2 x+13 y-15=0 \ldots( v )$
Eqs. (iv) and (v) are coincide
$\therefore \frac{2 g+8}{2}=\frac{2 f+2}{13}=\frac{c-16}{-15} $
$\Rightarrow 13 g-2 f=-50 \ldots $ (vi)
Solving Eqs. (iii) and (vi), we get
$g=\frac{11}{3} f=\frac{-7}{6}$
$\therefore $ Centre of circle,
$S=\left(\frac{-11}{3}, \frac{7}{6}\right)$