Let C_{1} be the curve obtained by the soluti | vedclass

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(B) For curve $C_{1}$: $\frac{dy}{dx} = \frac{y^{2} - x^{2}}{2xy}$.Put $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.$v + x\frac{dv}{dx} = \frac{

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$\frac{\pi}{2} - 1$

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(B) For curve $C_{1}$: $\frac{dy}{dx} = \frac{y^{2} - x^{2}}{2xy}$.Put $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.$v + x\frac{dv}{dx} = \frac{v^{2}x^{2} - x^{2}}{2x(vx)} = \frac{v^{2} - 1}{2v}$.$x\frac{dv}{dx} = \frac{v^{2} - 1}{2v} - v = \frac{v^{2} - 1 - 2v^{2}}{2v} = \frac{-(v^{2} + 1)}{2v}$.Separating variables: $\frac{2v}{v^{2} + 1} dv = -\frac{dx}{x}$.Integrating both sides: $\ln(v^{2} + 1) = -\ln x + C$.$\ln(\frac{y^{2}}{x^{2}} + 1) = -\ln x + C \Rightarrow \ln(\frac{x^{2} + y^{2}}{x^{2}}) + \ln x = C \Rightarrow \ln(\frac{x^{2} + y^{2}}{x}) = C$.Since it passes through $(1, 1)$,$\ln(\frac{1+1}{1}) = C \Rightarrow C = \ln 2$.So,$\frac{x^{2} + y^{2}}{x} = 2 \Rightarrow x^{2} + y^{2} - 2x = 0$ (Circle with center $(1, 0)$ and radius $1$).For curve $C_{2}$: $\frac{dy}{dx} = \frac{2xy}{x^{2} - y^{2}}$.This is the reciprocal of the slope of $C_{1}$ with a sign change,implying orthogonality or symmetry. Solving similarly with $y=vx$ leads to $x^{2} + y^{2} - 2y = 0$ (Circle with center $(0, 1)$ and radius $1$).The area enclosed by these two circles is $2 \int_{0}^{1} (\sqrt{1 - (x-1)^{2}} - x) dx = 2 [\frac{\pi}{4} - \frac{1}{2}] = \frac{\pi}{2} - 1$.

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