The differential equation of family of circles whose centre lie on \(\mathrm{X}\)-axis is

Let \((h, 0)\) be the centre of the circle and ' \(r\) ' be the radius.
\(
\therefore(\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-0)^2=\mathrm{r}^2 \Rightarrow(\mathrm{x}-\mathrm{h})^2+\mathrm{y}^2=\mathrm{r}^2
\)
Differentiating w.r.t. \(\mathrm{x}\), we get
\(
\therefore 2(\mathrm{x}-\mathrm{h})(1)+2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=0 \quad \Rightarrow \mathrm{h}=\mathrm{x}+\mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}
\)
Substituting value of ' \(h\) ' in eq. (1), we get
\(
y^2\left(\frac{d y}{d x}\right)^2+y^2=r^2 \Rightarrow y^2\left[1+\left(\frac{d y}{d x}\right)^2\right]=r^2
\)
Differentiating w.r.t. \(\mathrm{x}\), we get
\(
\begin{aligned}
& y^2\left(2 \frac{d y}{d x} \cdot \frac{d^2 y}{d x^2}\right)+\left[1+\left(\frac{d y}{d x}\right)^2\right] 2 y \frac{d y}{d x}=0 \\
& \therefore y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2+1=0
\end{aligned}
\)