If $Q$ is the inverse point of the point $\mathrm{P}(2,3)$ with respect to the circle $x^2+y^2-2 x-2 y+1=0$, then the circle with $\mathrm{PQ}$ as diameter is

• A

  $3 x^2+3 y^2-14 x-16 y+37=0$
• B

  $x^2+y^2-4 x-6 y+13=0$
• Correct answer
  C

  $5 x^2+5 y^2-16 x-22 y+33=0$
• D

  $2 x^2+2 y^2-3 x-3 y-11=0$

Solution

Let the co-ordinate of the point $Q$ is $(h, k)$. Equation of line $O P$ : $(y-3)=\frac{2}{1}(x-1)$ $\Rightarrow \quad 2 x-y=1$ ...(1) $\because(h, k)$ is on the line $O P$ $\therefore \quad 2 h-k=1 \Rightarrow k=2 h-1$
So, co-ordinate of $P$ is $(h, 2 h-1)$. Also, $O P \times O Q=r^2$ $\begin{aligned} & \Rightarrow \sqrt{1+4} \times O Q^{\prime}=r^2 \Rightarrow O Q=\frac{1}{\sqrt{5}} \\ & \Rightarrow \sqrt{(h-1)^2+(2 h-1-1)^2}=\frac{1}{\sqrt{5}}\end{aligned}$ $\begin{aligned} & \Rightarrow \quad 25 h^2-50 h+24=0 \\ & \Rightarrow \quad h=\frac{6}{5}, \frac{4}{5}\end{aligned}$ Let $h=\frac{6}{5}$. Then $k=2 \times \frac{6}{5}-1=\frac{7}{5}$ Then $Q=\left(\frac{6}{5}, \frac{7}{5}\right)$. Eqn. of line with $P Q$ as diameter is $\begin{aligned} & (x-2)\left(x-\frac{6}{5}\right)+(y-3)\left(y-\frac{7}{5}\right)=0 \\ & \Rightarrow \quad(x-2)(5 x-6)+(5 y-7)(y-3)=0 \\ & \Rightarrow 5 x^2+5 y^2-6 x-22 y+33=0 .\end{aligned}$

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