A) \[x\sqrt{1-{{y}^{2}}}-y\sqrt{1-{{x}^{2}}}=c\]
B) \[x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}}=c\]
C) \[x\sqrt{1+{{y}^{2}}}+y\sqrt{1+{{x}^{2}}}=c\]
D) None of these
Correct Answer: B
Solution :
\[\frac{dy}{dx}+\sqrt{\frac{1-{{y}^{2}}}{1-{{x}^{2}}}}=0\] Þ \[\int_{{}}^{{}}{\frac{dy}{\sqrt{1-{{y}^{2}}}}}=-\int_{{}}^{{}}{\frac{dx}{\sqrt{1-{{x}^{2}}}}}\] Þ \[{{\sin }^{-1}}y=-{{\sin }^{-1}}x+{{\sin }^{-1}}c\] Þ \[{{\sin }^{-1}}\left[ x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}} \right]={{\sin }^{-1}}c\] Þ \[x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}}=c\].
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