All circles passing through origin and having centre on $X-axis$ say at $(a,0)$ will have $ radius= a$

$\therefore $ General equation of such circles is

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

This equation has only one arbitrary constant.$a$.

$\therefore$ This equation can be differentiated only once to remove the arbitrary constant $a$

Differentiating (i) on both the sides we get

$2(x-a)+2y.\large\frac{dy}{dx}=0$........(ii)

$\Rightarrow\:x-a=-y.\large\frac{dy}{dx}$

and

$a=x+y\large\frac{dy}{dx}$

Substituting the values of $(x-a) \:\:and\:\: a$ in (i) we get

$y^2.(\large\frac{dy}{dx})^2$$+y^2=(x+y\large\frac{dy}{dx})^2$

$\Rightarrow\:y^2.(\large\frac{dy}{dx})^2$$+y^2=x^2+y^2.(\large\frac{dy}{dx})^2$$+2xy.\large\frac{dy}{dx}$

$\Rightarrow\:y^2=x^2+2xy.\large\frac{dy}{dx}$

is the differential equation of the given circles.