This question can be solved by using homogeneous equation concept.
Rearranging the equation, $\frac{dx}{dy}=\frac{x^{2}-y^{2}}{2xy}$ On RHS ,divide the numerator and denominator by $y^{2}$ and let $x=vy$.
So, $\frac{dx}{dy}=v+y\frac{dv}{dy}$ $=$ $\frac{v^{2}-1}{2v}$
$\Rightarrow$ $y\frac{dv}{dy}=\frac{-\left ( 1+v^{2} \right )}{2v}$
$\Rightarrow$ $\int \frac{2v}{1+v^{2}}dv=-\int \frac{dy}{y}$
$\Rightarrow$ $ln_{e}\left ( 1+v^{2} \right )=-ln_{e}\left ( y \right )+ln_{e}\left ( c \right )$
$\Rightarrow$ $ln_{e}\left ( \left ( 1+v^{2} \right )\times y \right )=ln_{e}\left ( c \right )$
$\Rightarrow$ $x^{2}+y^{2}=cy$ ---- $\left \{ x=vy \right \}$ substitution in the previous step
Using the given conditions i.e. $y=10$ for $x=0$, we get $c=10$
$\therefore$ $x^{2}+y^{2}=10y$
Rearranging and simplifying the above equation, we get $x^{2}+\left ( y-5 \right )^{2}=25$
Option A is the correct answer.