For the differential equation whose solution is $(\mathrm{x}-\mathrm{h}{)}^2+(\mathrm{y}-\mathrm{k}{)}^2={\mathrm{a}}^2$(a is a constant), its

We have $(\mathrm{x}-\mathrm{h}{)}^2+(\mathrm{y}-\mathrm{k}{)}^2={\mathrm{a}}^2$                (1)
Differentiating w.r.t.x, get
$2(\mathrm{x}-\mathrm{h})+2(\mathrm{y}-\mathrm{k})\frac{\mathrm{dy}}{\mathrm{dx}}=0$                          (2)
Differentiating w.r.t. x, we get
$1+{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2+(\mathrm{y}-\mathrm{k})\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=0$                      (3)
From equation (3),
$\mathrm{y}-\mathrm{k}=-\left(\frac{1+{\mathrm{p}}^2}{\mathrm{q}}\right),$where $\mathrm{p}=\frac{\mathrm{dy}}{\mathrm{dx}},\mathrm{q}=\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}$
Putting the value of $\mathrm{y}-\mathrm{k}$ in equation is (2),we get 
$\mathrm{x}-\mathrm{h}=\frac{\left(1+{\mathrm{p}}^2\right)\mathrm{p}}{\mathrm{q}}$
Substituting the values of $\mathrm{x}-\mathrm{h}$ and y - k in equation (1), we get
${\left(\frac{1+{\mathrm{p}}^2}{\mathrm{q}}\right)}^2\left(1+{\mathrm{p}}^2\right)={\mathrm{a}}^2$
or ${\left[1+{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2\right]}^3={\mathrm{a}}^2{\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)}^2$
which is the required differential equation.