Question

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

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

Answer 1

The correct option is B $y^2+6x=13$

Let $C(h,k)$ be centre (locus point)


$\therefore$ radius, $r=OC=\sqrt{h^2+k^2}$
$CM=|h-3|$
We know that,
$AB=2\sqrt{C{B}^2-C{M}^2}$
$\Rightarrow 4=2\sqrt{(h^2+k^2)-(h-3{)}^2}$
$\Rightarrow 4=(h^2+k^2)-(h^2-6h+9)$
$\Rightarrow k^2+6h=13$

$\therefore$Locus of $C(h,k)$ is
$y^2+6x=13$