A) \[({{x}^{2}}-{{y}^{2}})\frac{dy}{dx}+2xy=Q\]
B) \[({{x}^{2}}-{{y}^{2}})\frac{dy}{dx}=2xy\]
C) \[({{x}^{2}}-{{y}^{2}})\frac{dy}{dx}=xy\]
D) \[({{x}^{2}}-{{y}^{2}})\frac{dy}{dx}+xy=0\]
Correct Answer: B
Solution :
Equation of circle passing through origin and centre lies on y-axis \[{{(x-0)}^{2}}+\,\,{{(y-k)}^{2}}={{k}^{2}}\,\,\Rightarrow \,\,{{x}^{2}}-{{y}^{2}}+{{k}^{2}}-2ky={{k}^{2}}\] \[\Rightarrow \,\,\,{{x}^{2}}+{{y}^{2}}-2ky=0\] ... (i) Differentiate equation (i) w.r. to x \[2x+2y\frac{dy}{dx}-2k\frac{dy}{dx}=0\] \[K\,\frac{dy}{dx}=x+y\frac{dy}{dx}\Rightarrow \,\,k=\frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}}\] ? (ii) Put the value of k in (i) \[{{x}^{2}}+{{y}^{2}}-2\left( \frac{x+y\frac{dy}{dx}}{\frac{dy}{dx}} \right)\,\,y=0\] Solving this equation \[({{x}^{2}}-{{y}^{2}})\,\frac{dy}{dx}=2xy\]You need to login to perform this action.
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