III OBM

We have $(x-k{)}^2+x^2=1$, so $2x^2-2kx+k^2-1=0$. This has 0, 1 or 2 real solutions according as $k^2>2$, $k^2=2$ or $k^2<2$.

$k=\sqrt{2}$ gives $x=\frac{1}{\sqrt{2}}$, $y=\frac{1}{\sqrt{2}}$ or $y=-\frac{1}{\sqrt{2}}$, so there are two solutions to the original set. Similarly for $k=-\sqrt{2}$.

If $|k|<\sqrt{2}$, then $x=\frac{k\pm \sqrt{2-k^2}}{2}$, $y=\pm x$. That gives 4 solutions unless one of the values of $x$ is 0, in which case we get 3 solutions. So $k=1$ gives $(x,y)=(0,0)$ or $(1,1)$ or $(1,-1)$. $k=-1$ gives $(x,y)=(0,0)$ or $(-1,1)$ or $(-1,-1)$.

If $|k|>\sqrt{2}$, there are no solutions.