The curve amongst the family of curves represented by the differential equation $\left(x^2-y^2\right)dx+2xydy=0$ which passes through (1,1) is

The given differential equation is

$\begin{array}{l}\left(x^2-y^2\right)dx+2xydy=0\\ \Rightarrow \frac{dy}{dx}=\frac{y^2-x^2}{2xy}\end{array}$

This is a homogenous differential equation. Substituting

$y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$, we obtain

$\begin{array}{l}v+x\frac{dv}{dx}=\frac{v^2-1}{2v}\\ \Rightarrow x\frac{dv}{dx}=-\frac{v^2+1}{2v}\\ \Rightarrow \frac{2v}{v^2+1}dv=-\frac{dx}{x}\end{array}$

On integrating, we obtain

$\begin{array}{l}\log \left(v^2+1\right)=-\log x+\log C\\ \Rightarrow v^2+1=\frac{C}{x}\\ \Rightarrow x^2+y^2=Cx ... \left(i\right)\end{array}$

If (i) passes through (1, 1), then C = 2.

Putting C = 2 in (i), we obtain $x^2+y^2=2x$ which represents a circle with centre at (1,0) i.e. on x-axis.