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 :

Let $$y=y(x)$$ be the solution of the differential equation $$\left(2 x \log _e x\right) \frac{d y}{d x}+2 y=\frac{3}{x} \log _e x, x>0$$ and $$y\left(e^{-1}\right)=0$$. Then, $$y(e)$$ is equal to

Let $$y=y(x)$$ be the solution of the differential equation $$\left(1+x^2\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}$$, $$y(1)=0$$. Then $$y(0)$$ is

The differential equation of the family of circles passing through the origin and having centre at the line $$y=x$$ is :