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

We have ${(x-h)}^2+{(y-k)}^2=a^2$

 Differentiating w.r.t. x, we get

  $\begin{array}{l}2(x-h)+2(y-k)\frac{dy}{dx}=0\\ or(x-h)+(y-k)\frac{dy}{dx}=0\end{array}$

 Differentiating w.r.t. x, we get

   $1+{\left(\frac{dy}{dx}\right)}^2+(y-k)\frac{d^{2}y}{d{x}^2}=0$

From equation (3),

$y-k=-\left(\frac{1+p^2}{q}\right),wherep=\frac{dy}{dx},q=\frac{d^{2}y}{d{x}^2}$

Putting the value of y-k in equation (2), we get

  $x-h=\frac{(1+p^2)p}{q}$

  Substituting the values of x-h and y-k in equation (1) , we get

$\begin{array}{l}{\left(\frac{1+p^2}{q}\right)}^2(1+p^2)=a^2\\ or{[1+{\left(\frac{dy}{dx}\right)}^2]}^3=a^2{\left(\frac{d^{2}y}{d{x}^2}\right)}^2\end{array}$

 Which is the required differential equation.