Find the locus of a point which moves such that $k$ times its distance from the $Y$ axis is equal to the square of its distance from the origin.
Find the center and the radius of the resulting locus.
Let $P \left(x, y\right)$ be a point on the required locus.
Distance of point $P$ from the $Y$ axis is $= x$
Distance of point $P$ from the origin $= \sqrt{\left(x - 0\right)^2 + \left(y - 0\right)^2} = \sqrt{x^2 + y^2}$
As per question,
$k x = \left(\sqrt{x^2 + y^2}\right)^2$
i.e. $\;$ $kx = x^2 + y^2$
i.e. $\;$ $x^2 + y^2 - kx = 0$
is the equation of the required locus which is the equation of a circle.
Comparing with the standard equation of a circle: $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
$2g = -k$ $\implies$ $g = \dfrac{-k}{2}$, $\;$ $f = 0$, $\;$ $c = 0$
Center of the circle $= \left(-g, -f\right) = \left(\dfrac{k}{2}, 0\right)$
Radius of the circle $= \sqrt{g^2 + f^2 - c} = \sqrt{\left(\dfrac{-k}{2}\right)^2 + 0 + 0} = \sqrt{\dfrac{k^2}{4}} = \dfrac{k}{2}$