JEE Math Practice Question

Let the solution curve of the differential equation $$x \mathrm{~d} y=\left(\sqrt{x^{2}+y^{2}}+y\right) \mathrm{d} x, x>0$$, intersect the line $$x=1$$ at $$y=0$$ and the line $$x=2$$ at $$y=\alpha$$. Then the value of $$\alpha$$ is :

1. A.$$\frac{1}{2}$$
2. B.$$\frac{3}{2}$$
3. C.$$-$$$$\frac{3}{2}$$
4. D.$$\frac{5}{2}$$