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

Taking $\mathrm{x}=\mathrm{rcos}\mathrm{\theta } \mathrm{\&} \mathrm{y}=\mathrm{rsin}\mathrm{\theta }\text{ so that }$

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

We have $\mathrm{xdx}+\mathrm{ydy}=\mathrm{r}\cdot \mathrm{dr}\text{ and }\mathrm{xdy}-\mathrm{ydx}={\mathrm{x}}^2\mathrm{se}{c}^2\mathrm{\theta d\theta }={\mathrm{r}}^2\mathrm{d\theta }$

The given equation can be transformed into $\frac{\mathrm{rdr}}{{\mathrm{r}}^2\mathrm{d\theta }}=\sqrt{\left(\frac{{\mathrm{a}}^2-{\mathrm{r}}^2}{{\mathrm{r}}^2}\right)}$

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

Integrating both sides then we get

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

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