Official solution
(x2 - y2) dx + 2xydy = 0
\[ \Rightarrow \] \[{{dy} \over {dx}}\] = \[{{{y^2} - {x^2}} \over {2xy}}\]
Let y = vx
\[{{dy} \over {dx}}\] = v + x \[{{dv} \over {dx}}\]
\[ \Rightarrow \] v + x\[{{dv} \over {dx}}\] = \[{{{v^2}{x^2} - {x^2}} \over {2v{x^2}}}\] \[ \Rightarrow \] v + x\[{{dv} \over {dx}}\] = \[{{{v^2} - 1} \over {2v}}\]
\[ \Rightarrow \] x\[{{dv} \over {dx}}\] = \[{{ - {v^2} - 1} \over {2v}}\]
\[ \Rightarrow \] \[{{2vdv} \over {{v^2} + 1}}\] = - \[{{dx} \over x}\]
After intergrating, we get
\[\ln \left| {{v^2} + 1} \right|\] = - ln\[\left| x \right|\] + lnc
\[{{y{}^2} \over {{x^2}}}\] + 1 = \[{c \over x}\]
As curve passes through the point (1, 1), so 1 + 1 = c
\[ \Rightarrow \] c = 2
x2 + y2 - 2x = 0, which is a circle of radius one.