What is the equation of the circle which passes through the points $(3,-2)$ and $(-2,0)$ and having its centre on the line $2 x-y-3=0$ ?

• A

  $x^2+y^2+3 x+2=0$
• Correct answer
  B

  $x^2+y^2+3 x+12 y+2=0$
• C

  $x^2+y^2+2 x=0$
• D

  $x^2+y^2=5$

Solution

Given, centre of circle lies on line $2 x-y-3=0$ Let $\mathrm{x}=\mathrm{h}$ $2 h-y-3=0 \Rightarrow y=2 n-3$
$\therefore$ centre of circle $=(h, 2 h-3)$ We know, $\mathrm{OA}=\mathrm{OB}$ $\begin{aligned} & \Rightarrow(\mathrm{h}-3)^2+(2 \mathrm{~h}-3+2)^2=(\mathrm{h}+2)^2+(2 \mathrm{~h}-3)^2 \\ & \Rightarrow \mathrm{~h}^2-6 \mathrm{~h}+9+(2 \mathrm{~h}-1)^2=\mathrm{h}^2+4 \mathrm{~h}+4+4 \mathrm{~h}^2-12 \mathrm{~h}+9 \\ & \Rightarrow \mathrm{~h}^2-6 \mathrm{~h}+9+4 \mathrm{~h}^2-4 \mathrm{~h}+1 \\ & =\mathrm{h}^2+4 \mathrm{~h}+4+4 \mathrm{~h}^2-12 \mathrm{~h}+9 \\ & \Rightarrow-10 \mathrm{~h}+10=-8 \mathrm{~h}+13 \\ & \Rightarrow-2 \mathrm{~h}=3 \Rightarrow \mathrm{~h}=\frac{-3}{2} \\ & \therefore \text { centre }=\left(\frac{-3}{2}, 2\left(\frac{-3}{2}\right)-3\right)=\left(\frac{-3}{2},-6\right) \\ & \therefore \text { radius }=(\mathrm{h}+2)^2+(2 \mathrm{~h}-3)^2 \\ & =\left(\frac{-3}{2}+2\right)^2+\left[2\left(\frac{-3}{2}\right)-3\right]^2=\frac{1}{4}+36 \end{aligned}$ $\therefore$ Equation of circle $\begin{aligned} & \Rightarrow\left(\mathrm{x}+\frac{3}{2}\right)^2+(\mathrm{y}+6)^2=\frac{1}{4}+36 \\ & \Rightarrow \mathrm{x}^2+\mathrm{y}^2+3 \mathrm{x}+12 \mathrm{y}+2=0 \end{aligned}$

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