The differential equation of all circles which pass through the origin and whose centre lies on y-axis, is

$\left(x^2-y^2\right) \frac{d y}{d x}-2 x y=0$

The equation of the family of circles passing through the origin and having centres on $y$-axis is

$x^2+(y-a)^2=a^2 \Rightarrow x^2+y^2-2 a y=0$       .....(i)

Differentiating w.r.t. $x$, we get

$2 x+2 y \frac{d y}{d x}-2 a \frac{d y}{d x}=0 \Rightarrow a=\frac{x+y \frac{d y}{d x}}{\frac{d y}{d x}}=x \frac{d x}{d y}+y$

Substituting the value of $a$ in (i), we obtain

$x^2+y^2-2 y\left(x \frac{d x}{d y}+y\right)=0$

$x^2+y^2-2 y\left(x \frac{d x}{d y}+y\right)=0$

$\Rightarrow x^2-y^2-2 x y \frac{d x}{d y}=0 \Rightarrow\left(x^2-y^2\right) \frac{d y}{d x}-2 x y=0$