Suppose the solution of the differential equation $\frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-2…
Suppose the solution of the differential equation $\frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-2 \alpha y-(\beta \gamma-4 \alpha)}$ represents a circle passing through origin. Then the radius of this circle is :
2
$\sqrt{17}$
$\frac{1}{2}$
$\frac{\sqrt{17}}{2}$
Solution
$\begin{aligned} & \frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-y(2 \alpha+\beta)+4 \alpha} \\ & \beta x d y-(2 \alpha+\beta) y d y+4 \alpha d y=(2+\alpha) x d x-\beta y d x+2 d x \\ & \beta(x d y+y d x)-(2 \alpha+\beta) y d y+4 \alpha d y=(2+\alpha) x d x+2 d x \\ & \beta x y-\frac{(2 \alpha+\beta) y^2}{2}+4 \alpha y=\frac{(2+\alpha) x^2}{2}\end{aligned}$
$\begin{aligned} & \Rightarrow \beta=0 \text { for this to be circle } \\ & (2+\alpha) \frac{x^2}{2}+\alpha y^2+2 x-4 \alpha y=0\end{aligned}$
$\begin{aligned} & \text { i.e. } 2 x^2+2 y^2+2 x-8 y=0 \\ & x^2+y^2+x-4 y=0 \\ & r d=\sqrt{\frac{1}{4}+4}=\frac{\sqrt{17}}{2}\end{aligned}$
