The differential equation of all circles which pass through the origin and whose centres lie on \(\mathrm{Y}\)-axis is

Circle passes through origin and centre lie on \(\mathrm{Y}\)-axis.
Let \((0, k)\) be centre and ' \(k\) ' be radius
\(\therefore \quad\) Equation of circle is
\(\begin{aligned}
& (x-0)^2+(y-\mathrm{k})^2=\mathrm{k}^2 \\
& x^2+y^2-2 y \mathrm{k}+\mathrm{k}^2=\mathrm{k}^2 \\
& x^2+y^2-2 \mathrm{k} y=0 \\
& x^2+y^2=2 \mathrm{k} y ...(i)\\
& \frac{x^2+y^2}{2 y}=\mathrm{k} ...(ii)
\end{aligned}\)
Differentiating equation (i) with respect to \(x\), we get
\(\begin{aligned}
& 2 x+2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=2 \mathrm{k} \frac{\mathrm{d} y}{\mathrm{~d} x} \\
& 2 x+2 y \frac{\mathrm{d} y}{\mathrm{~d} x}-2 \mathrm{k} \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\
& 2 x+2(y-\mathrm{k}) \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\
& 2 x+2\left[y-\left(\frac{x^2+y^2}{2 y}\right)\right] \frac{\mathrm{d} y}{\mathrm{~d} x}=0 ...[From(ii)]\\
& 2 x+2\left[\frac{2 y^2-x^2-y^2}{2 y}\right] \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\
& 2 x+\left(\frac{y^2-x^2}{y}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\
& 2 x y+\left(y^2-x^2\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=0 \\
& \text { i.e. }\left(x^2-y^2\right) \frac{\mathrm{d} y}{\mathrm{~d} x}-2 x y=0
\end{aligned}\)