A) A circle of radius two
B) A circle of radius one
C) A hyperbola
D) An ellipse
Correct Answer: B
Solution :
\[({{x}^{2}}-{{y}^{2}})dx+2xydy=0\] \[\Rightarrow \frac{dy}{dx}=\frac{{{y}^{2}}-{{x}^{2}}}{2xy}\] Put \[y=vx\] \[\frac{dy}{dx}=v+x\frac{dv}{dx}\] \[\Rightarrow v+x\frac{dv}{dx}=\frac{{{v}^{2}}{{x}^{2}}-{{x}^{2}}}{2v{{x}^{2}}}\] \[\Rightarrow v+x\frac{dv}{dx}=\frac{{{v}^{2}}-1}{2v}\] \[\Rightarrow x\frac{dv}{dx}=\frac{-{{v}^{2}}-1}{2v}\] \[\Rightarrow \frac{2vdv}{{{v}^{2}}+1}=-\frac{dx}{x}\] Integrating we get; \[\ln |{{v}^{2}}+1|=-\ln |x|+lnc\] \[\frac{{{y}^{2}}}{{{x}^{2}}}+1=\frac{c}{x}\] Putting (1,1) \[c=2\] \[{{x}^{2}}+{{y}^{2}}-2x=0\] hence its is a circle of radius 1
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