This is really just a comment, not a complete answer.
Suppose $r$ is integral. The distance from the point $[1,r]$ to $C(r)$ is $$\sqrt{1+r^2}-r$$ which has asymptotics $\frac{1}{2r}$ and hence for integral diameters the answers are affirmative.
If $r$ is nonintegral, then the point $[0,\lceil r \rceil]$ is closer to $C(r)$ than $[0,r]$ and its distance to $C(r)$ $$ \lceil r \rceil - r $$ gets arbitrarily close to 1 (as Abhinav Kumar pointed out in the comments). So for nonintegral diameters one really has to pick a good lattice point inside the first quadrant.