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 :

<p>$$\begin{aligned} & \frac{d y}{d x}=\frac{(2+\alpha) x-\beta y+2}{\beta x-2 \alpha y-(\beta \gamma-4 \alpha)} \\ & \beta x d y-2 \alpha y d y-(\beta \gamma-4 \alpha) d y \\ & =2 x d x+\alpha x d x-\beta y d x+2 d x \\ & \beta \int(x d y+y d y)-\alpha y^2-(\beta \gamma-4 x) y=x^2+\frac{\alpha x^2}{2}+2 x \\ & \beta x y-\alpha y^2-(\beta \gamma-4 \alpha) y=x^2+\frac{\alpha x^2}{2}+2 x \\ & \left(1+\frac{\alpha}{2}\right) x^2+\alpha y^2-\beta x y+2 x+(\beta \gamma-4 \alpha) y=0 \\ & \because \text { this represents circle passing through origin } \\ & \Rightarrow \beta=0 \text { and } 1+\frac{\alpha}{2}=\alpha \\ & \Rightarrow \alpha=2 \end{aligned}$$</p> <p>$$\begin{aligned} & \therefore C: 2 x^2+2 y^2+2 x-8 y=0 \\ & x^2+y^2+x-4 y=0 \\ & \text { Radius }=\sqrt{\frac{1}{4}+4-0} \\ & \quad=\frac{\sqrt{17}}{2} \end{aligned}$$</p>