A curve passes through (1,0) and satisfies xdy/dx=y+√(x^2+y^2 for x>0. Which of the following statements is correct?

The equation is homogeneous: put $y=vx$ so that $v+x\,v'=v+\sqrt{1+v^2}$, i.e. $\frac{dv}{\sqrt{1+v^2}}=\frac{dx}x$. Integrating, $\sinh^{-1}v=\ln x+c$, so $v=\sinh(\ln x+c)=\frac12\left(Cx-\frac1{Cx}\right)$ with $C=e^c>0$ and $y=\frac12\left(Cx^2-\frac1C\right)$. Passing through $(1,0)$ forces $C=1$, giving $y=\frac{x^2-1}{2}$ (A correct; direct substitution confirms $xy'=x^2=y+\sqrt{x^2+y^2}$). Every member has vertex $\left(0,-\frac1{2C}\right)$ on the $y$-axis, not the $x$-axis (B incorrect). The area is $\int_1^3\frac{x^2-1}2dx=\left[\frac{x^3}6-\frac x2\right]_1^3=\frac{10}3\neq\frac83$ (C incorrect). The slope at $(1,0)$ is $y'(1)=1$, so the tangent $y=x-1$ meets the $y$-axis at $(0,-1)$, not $(0,1)$ (D incorrect). Answer: A only.