Let $(\mathrm{h}, 8)$ be the centre of the circle.
Since circle touches $X$ axis, radius $=8$
$\therefore(x-h)^{2}+(y-8)^{2}=(8)^{2}$...(1)
Differentiating w.r.t. x, we get
$\begin{aligned}
& 2(x-h)+2(y-8) \frac{d y}{d x}=0 \\
\therefore \quad &(x-h)=-(y-8) \frac{d y}{d x}
\end{aligned}$
Substituting value of $(x-h)$ in eq. (1), we get
$\begin{aligned}
&\left[-(y-8) \frac{d y}{d x}\right]^{2}+(y-8)^{2}=64 \\
\therefore &(y-8)^{2}\left[\left(\frac{d y}{d x}\right)^{2}+1\right]=64
\end{aligned}$
