Suppose the solution of the differential equation $$\frac{dy}{dx} = \frac{(2+\alpha)x - \beta y + 2}{\beta x - 2\alpha y - (\beta\gamma - 4\alpha)}$$ represents a circle passing through origin. Then the radius of this circle is :

We rewrite the given differential equation in differential form as $$\frac{dy}{dx} \;=\;\frac{(2+\alpha)x \;-\;\beta y \;+\;2}{\beta x \;-\;2\alpha y \;-\;(\beta\gamma -4\alpha)}.$$ Putting all terms on one side gives $$\bigl[-\bigl((2+\alpha)x -\beta y +2\bigr)\bigr]\,dx \;+\;\bigl[\beta x -2\alpha y -(\beta\gamma -4\alpha)\bigr]\,dy \;=\;0.$$ Thus we identify $$M(x,y) \;=\;-\,\bigl((2+\alpha)x -\beta y +2\bigr),\quad N(x,y) \;=\;\beta x -2\alpha y -(\beta\gamma -4\alpha).$$

We check for exactness. A differential form $$M\,dx + N\,dy=0$$ is exact if $$\frac{\partial M}{\partial y} \;=\;\frac{\partial N}{\partial x}.$$ Here $$\frac{\partial M}{\partial y} = -(-\beta)=\beta,\qquad \frac{\partial N}{\partial x} = \beta.$$ Since these are equal, the equation is exact.

Because it is exact, there exists a potential function $$\phi(x,y)$$ such that $$\frac{\partial \phi}{\partial x}=M,\quad \frac{\partial \phi}{\partial y}=N.$$ We find $$\phi$$ by integrating $$M$$ with respect to $$x$$: $$\phi(x,y)=\int M\,dx =\int -\bigl((2+\alpha)x -\beta y +2\bigr)\,dx =-\frac{(2+\alpha)x^2}{2}+\beta x y -2x +h(y),$$ where $$h(y)$$ is an unknown function of $$y$$.

Next we impose $$\displaystyle \frac{\partial \phi}{\partial y}=N$$. From our $$\phi$$ we get $$\frac{\partial \phi}{\partial y} =\beta x +h'(y) \quad\stackrel{!}{=} \; \beta x -2\alpha y -(\beta\gamma -4\alpha).$$ Hence $$h'(y) = -2\alpha y -(\beta\gamma -4\alpha).$$ Integrating with respect to $$y$$ gives $$h(y) = -\alpha y^2 -(\beta\gamma -4\alpha)\,y +C_1.$$

Thus the general solution $$\phi(x,y)=C$$ is $$-\frac{(2+\alpha)x^2}{2}+\beta x y -2x \;-\;\alpha y^2 \;-\;(\beta\gamma -4\alpha)y +C_1 =0.$$ Multiplying by $$-2$$ and renaming the constant yields the family of curves $$ (2+\alpha)x^2 \;-\;2\beta x y \;+\;2\alpha y^2 \;+\;4x \;+\;2\beta\gamma\,y \;-\;8\alpha\,y +C_2 \;=\;0. $$

For this family to represent a circle, two conditions must hold: Case 1: the coefficient of $$xy$$ must be zero, yielding $$-2\beta =0\;\Longrightarrow\;\beta=0.$$ Case 2: the coefficients of $$x^2$$ and $$y^2$$ must be equal, giving $$2+\alpha =2\alpha\;\Longrightarrow\;\alpha=2.$$

Substituting $$\alpha=2$$ and $$\beta=0$$ into the equation, we get $$4x^2 +4y^2 +4x -16y +C_2 =0.$$ Dividing by 4 simplifies to $$x^2 +y^2 +x -4y +C_3 =0.$$

We now use the condition that the circle passes through the origin $$(0,0)$$. Substituting $$(x,y)=(0,0)$$ gives $$0^2 +0^2 +0 -0 +C_3 =0\;\Longrightarrow\;C_3=0.$$ Hence the specific circle is $$x^2 +y^2 +x -4y =0.$$

We complete the square to find its centre and radius: $$x^2 +x +y^2 -4y =0 \;\Longrightarrow\;(x+\tfrac12)^2 -\tfrac14 + (y-2)^2 -4 =0$$ $$\Longrightarrow (x+\tfrac12)^2 + (y-2)^2 = \tfrac14 +4 = \tfrac{17}{4}.$$ Therefore the radius is $$\boxed{\frac{\sqrt{17}}{2}}.$$