We are asked to form the differential equations of all circles which pass through the orgin and whose centers lies on y-axis
Since it is given that the center lies on the y-axis, the sketch of the circle is as shown
The equation of the family of circles touching the x-axis at the origin is
$(x-0)^2+(y-a)^2=a^2$
=>$x^2+y^2-2ay=0$------(1)
Where a is a parameter
This equation has only one arbitary constant, hence let us differentiate it once w.r.t $x$
$2x+2y.\large\frac{dy}{dx}$$-2a \large\frac{dy}{dx}=$$0$
$=>a=\large\frac{x+y(dy/dx)}{dy/dx}$
Substituting this value of a in equ(1) we get,
$x^2+y^2=2y \bigg(\large\frac{x+y(dy/dx)}{dy/dx}\bigg)$
On simplification we get
=>$ (x^2-y^2)\large\frac{dy}{dx}$$=2xy$
or $ (x^2-y^2)\large\frac{dy}{dx}$$-2xy=0$ is the required equation