question_answer

The differential equation representing a family of circles touching the y-axis at the origin is :

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.