The curve amongst the family of curves, represented

$\left(x^{2}-y^{2}\right) d x+2 x y d y=0$

$\frac{d y}{d x}=\frac{y^{2}-x^{2}}{2 x y}$

Put $\quad y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Solving we get,

$\int \frac{2 v}{v^{2}+1} d v=\int-\frac{d x}{x}$

$\ln \left(v^{2}+1\right)=-\ln x+C$

$\left(y^{2}+x^{2}\right)=C x$

$1+1=C \Rightarrow C=2$

$y^{2}+x^{2}=2 x$

$\therefore$ Option (2)