If two circles $x^2+y^2-6 x-6 y+13=0$ and $x^2+y^2-8 y$ $+9=0$ intersect at $\mathrm{A}$ and $\mathrm{B}$, then the focus of the parabola whose directrix is the line $A B$ and vertex is the point $O$ $(0,0)$ is

• A

  $\left(\frac{3}{5}, \frac{1}{5}\right)$
• Correct answer
  B

  $\left(-\frac{3}{5}, \frac{1}{5}\right)$
• C

  $\left(-\frac{3}{5}, -\frac{1}{5}\right)$
• D

  $\left(\frac{3}{5}, -\frac{1}{5}\right)$

Solution

$C_1: x^2+y^2-6 x-6 y+13=0$ $C_2: x^2+y^2-8 y+9=0$ Equation of common chord $A B \equiv C_1-C_2=0$ $\begin{aligned} & \therefore-6 x-6 y+13+8 y-9=0 \\ & \Rightarrow 6 x-2 y=4 \\ & \Rightarrow 3 x-y=2 \end{aligned}$ $\therefore \quad$ Directrix is $3 x-y-2=0 ... (i)$ $m_D=3$ $\therefore \quad$ Slope of axis $=\frac{-1}{3}$ Equation of axis $\Rightarrow y=\frac{-1}{3} x$ $\Rightarrow x+3 y=0... (ii)$ $\therefore \quad$ Foot of perpendicular on directrix is intersection of directrix and axis. $\therefore \quad$ From (i) and (ii), $x=\frac{3}{5}, y=\frac{-1}{5}$
$\begin{aligned} & \therefore \quad \frac{\alpha+\frac{3}{5}}{2}=0 \Rightarrow \alpha=\frac{-3}{5} \\ & \Rightarrow \frac{\beta-\frac{1}{5}}{2}=0 \Rightarrow \beta=\frac{1}{5}\end{aligned}$ $\therefore \quad$ Focus is $\left(\frac{-3}{5}, \frac{1}{5}\right)$.

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