The curve satisfying the differential equation (x²–y²) dx + 2xy dy = 0 and passes through point (1, 1) is a circle of area ....... .

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

$$\begin{gathered} \begin{array}{l}put y=vx \\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\\ \Rightarrow v+x\frac{dv}{dx}=\frac{v^2{x}^2-x^2}{2v{x}^2}\\ \Rightarrow v+x\frac{dv}{dx}=\frac{v^2-1}{2v}\\ \\ \Rightarrow x\frac{dv}{dx}=\frac{-v^2-1}{2v}\\ \Rightarrow \frac{2vdv}{v^2+1}=-\frac{dx}{x}\end{array} \end{gathered}$$

$\begin{array}{l}integrate,\\ \ln \left|{\mathrm{v}}^2+1\right|=-\ln |\mathrm{x}|+\ln c\\ \frac{{\mathrm{y}}^2}{{\mathrm{x}}^2}+1=\frac{\mathrm{c}}{\mathrm{x}}\end{array}$

passing through (1,1) then

$\begin{array}{l}c=2\\ x^2+y^2-2x=0\end{array}$

it is a circle having radius 1