The locus of centre of a circle which passes through the origin and cuts off a length of 4 unit from the line $x=3$ is

• A

  $y^2+6 x=0$
• Correct answer
  B

  $y^2+6 x=13$
• C

  $y^2+6 x=10$
• D

  $x^2+6 y=13$

Solution

Let centre of circle be $C(-g,-f)$, then equation of circle passing through origin be $x^2+y^2+2 g x+2 f y=0$
$\begin{array}{lc} \therefore \text { Distance, } d=|-g-3|=g+3 \\ \text { In } \triangle A B C, & (B C)^2=A C^2+B A^2 \\ \Rightarrow & g^2+f^2=(g+3)^2+2^2 \\ \Rightarrow & g^2+f^2=g^2+6 g+9+4 \\ \Rightarrow & f^2=6 g+13 \end{array}$ Hence, required locus is $y^2+6 x=13$

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