 Solution

Set $r=\sqrt{x^2+y^2}$. Wherever $r>0$, the chain rule and the given implicit differential equation giveThus $r=x+C$ and $C=r-x\ge 0$. Squaring without losing the square-root sign restriction giveswith either sign of $y$ on its interval of definition. These parabolas satisfy $x+C>0$. Their vertex $(-C/2,0)$ is a regular point as a parametrized curve but cannot belong to a differentiable graph $y(x)$ satisfying the equation: there $y{y}^{\prime }=0$ while $-x+r=C>0$. The degenerate case $C=0$ gives the separate solution $\boxed{y=0\text{ for }x\ge 0}$ in the differential-form interpretation printed in the paper. It would be lost by dividing the equation by $y$.