The differential equation of all circles which pass through the origi…

If (0, a) is centre on y-axis, then its radius is a because it passes through origin.

∴ Equation of circle is $x^2+(y-a)^2=a^2$

$\Rightarrow x^2+y^2-2 ay=0$                .......(1)

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

Using (1) in (2), $2 x+2 y \frac{d y}{d x}-\frac{x^2+y^2}{y} \frac{d y}{d x}=0$

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

$\Rightarrow \frac{d y}{d x}=\frac{2 x y}{x^2-y^2}$

Hence (1) is the correct answer.