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

Answer 1

Answer: (B)

Let centre of circle be $C(-g,-f)$, then equation of circle passing through origin be
$x^2+y^2+2,gx+2fy=0$
$\therefore$ Distance, $d=|-g-3|=g+3$
In $\Delta ABC,(BC)=A{C}^2+B{A}^2$
$\Rightarrow g^2+f^2=(g+3{)}^2+2^2$
$\Rightarrow g^2+f^2=g^2+6g+9+4$
$\Rightarrow f^2=6g+13$
Hence, required locus is $y^2+6x=13$