A) \[{{x}^{2}}+{{y}^{2}}-2xy\frac{dy}{dx}=0\]
B) \[{{x}^{2}}+{{y}^{2}}+2xy\frac{dy}{dx}=0\]
C) \[{{x}^{2}}-{{y}^{2}}-2xy\frac{dy}{dx}=0\]
D) \[{{x}^{2}}-{{y}^{2}}+2xy\frac{dy}{dx}=0\]
Correct Answer: D
Solution :
Since the circle touches the y-axis, therefore the centre lies on the x-axis. Let the centre be (h, 0) \[\Rightarrow \] radius of circle \[=h\] \[\therefore \] The equation of circle is given by \[{{(x-h)}^{2}}+{{(y-0)}^{2}}={{h}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-2hx=0\] ... (i) On differentiating both sides w.r.t. x, we get \[2x+2y\frac{dy}{dx}-2h=0\] \[\Rightarrow \] \[h=x+y\frac{dy}{dx}\] Putting the value of h in equation (i) \[{{x}^{2}}+{{y}^{2}}-2x\left( x+y\frac{dy}{dx} \right)=0\] \[\Rightarrow \] \[-{{x}^{2}}+{{y}^{2}}-2xy\frac{dy}{dx}=0\] \[\Rightarrow \] \[{{x}^{2}}-{{y}^{2}}+2xy\frac{dy}{dx}=0\] This is the required differential equation.You need to login to perform this action.
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