The solution of   is $\left(\frac{xdx+ydy}{xdy-ydx}\right)=\sqrt{\left(\frac{a^2-x^2-y^2}{x^2+y^2}\right)}$

Taking $x=r\cos \theta \&y=r\sin \theta$  so that 

$x^2+y^2=r^2\&\frac{y}{x}=\tan \theta$

We have $xdx+ydy=r·dr$  and  $xdy-ydx=x^2{\mathrm{sex}}^2\theta d\theta =r^{2}d\theta$

The given equation can be transformed into 

$\frac{rdr}{r^{2}d\theta }=\sqrt{\left(\frac{a^2-r^2}{r^2}\right)}$$\Rightarrow \frac{dr}{d\theta }=\sqrt{(a^2-r^2)}\Rightarrow \frac{dr}{\sqrt{(a^2-r^2)}}=d\theta$

Integrating both sides then we get

${\sin }^{-1}\left(\frac{r}{a}\right)=\theta +C$

$\Rightarrow \sqrt{x^2+y^2}=a\sin \left[{\tan }^{-1}(y/x)+\text{constant}\right]$