Question

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. $y^{2}+6 x=0$
• B. $y^{2}+6 x=13$
• C. $y^{2}+6 x=10$
• D. $x^{2}+6 y=13$

Answer 1

Answer: (B)

Let centre of circle be $C(-g,-f)$, then equation of circle passing through origin be
$x^{2}+y^{2}+2 g x+2 f y=0$

$\therefore $ Distance, $d=|-g-3|=g+3$
In $\Delta A B C,(B C)^{2}=A C^{2}+B A^{2}$
$\Rightarrow g^{2}+f^{2}=(g+3)^{2}+2^{2}$
$\Rightarrow g^{2}+f^{2}=g^{2}+6 g+9+4$
$\Rightarrow f^{2}=6 g+13$
Hence, required locus is $y^{2}+6 x=13$