A) \[{{y}^{2}}={{x}^{2}}+2xy\,\frac{dy}{dx}\]
B) \[{{y}^{2}}={{x}^{2}}-2xy\,\frac{dy}{dx}\]
C) \[{{x}^{2}}={{y}^{2}}+xy\,\frac{dy}{dx}\]
D) \[{{x}^{2}}={{y}^{2}}+3xy\,\frac{dy}{dx}\]
Correct Answer: A
Solution :
The system of circles passing through origin and centre lies on x-axis is \[{{x}^{2}}+{{y}^{2}}-2hx=0\] ?..(i) On differentiating w. r. t. x, we get \[2x+2y\frac{dy}{dx}-2h=0\] \[\Rightarrow \] \[2x+2y\frac{dy}{dx}-\left( \frac{{{x}^{2}}+{{y}^{2}}}{x} \right)=0\] [from Eq. (i)] \[\Rightarrow \] \[{{x}^{2}}-{{y}^{2}}+2xy\frac{dy}{dx}=0\]
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