The general equation of a circle having centre at $\left ( h,k \right )$ and radius $r$ is : $\left ( x-h \right )^{2}+\left ( y-k \right )^{2}=r^{2}$
Now, if the circle touches $y-axis$ at the origin, it's Centre would be lying at point $\left ( a,0 \right )$ ($"a"$ is assumed) and radius $a$ units.
So, the equation of circle would be: $\left ( x-a \right )^{2}+\left ( y-0 \right )^{2}=a^{2}$
$\Rightarrow$ $x^{2}+y^{2}=2ax$ ----- $\left ( 1 \right )$
Differentiating on both sides w.r.t $x$, we get
$2x+2y\frac{dy}{dx}=2k$ $\Rightarrow$ $a=x+y\frac{dy}{dx}$
Now, substituting the value of $a$ in $\left ( 1 \right )$, we get
$x^{2}+y^{2}=2x \left ( x+y\frac{dy}{dx} \right )$
Rearranging the terms and simplifying, we get the differential equation as: $x^{2}-y^{2}+2xy\frac{dy}{dx}=0$
Option D is the correct answer.
