The curve satisfying the differential equation, $(x^{2} - y^{2}) d x + 2 x y d y = 0$ and passing through the point $(1, 1)$ is :

a circle of radius one

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a circle of radius two

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an ellipse

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a hyperbola

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Solution

The correct option is A a circle of radius one
$(x^{2} - y^{2}) d x + 2 x y d y = 0$
$\Rightarrow \frac{d y}{d x} = \frac{1}{2} \frac{y}{x} (1 - {(\frac{x}{y})}^{2})$
Put $\frac{y}{x} = v \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x}$
$\Rightarrow v + x \frac{d v}{d x} = \frac{v}{2} (1 - \frac{1}{v^{2}})$
$\Rightarrow \frac{2 v}{v^{2} + 1} d v = - \frac{d x}{x}$
Integrating both side
$\Rightarrow \int \frac{2 v}{v^{2} + 1} d v = - \int \frac{d x}{x}$
$\Rightarrow ln (v^{2} + 1) = - ln x + ln c$
$\Rightarrow \frac{y^{2} + x^{2}}{x^{2}} = \frac{c}{x}$
$\Rightarrow c = \frac{y^{2} + x^{2}}{x}$
At $(1, 1), c = 2$
$\Rightarrow x^{2} + y^{2} - 2 x = 0$
$∴$ radius $r = 1$

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