I hope I'm not making some stupid mistake. Suppose $r$ is integral. The distance from the point $[1,r]$ to $C(r)$ is $\sqrt(1+r^2)-r$ which goes to zero as $r$ goes to infinity. For nonintegral $r$ the point $[0,\lceil r \rceil]$ is even closer to $C(r)$ and hence we have an upper bound $\beta(r) < \frac{1}{2r}$ for sufficiently big $r$.

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.

I hope I'm not making some stupid mistake. Suppose $r$ is integral. The distance from the point $[1,r]$ to $C(r)$ is $\sqrt(1+r^2)-r$ which goes to zero as $r$ goes to infinity. For nonintegral $r$ the point $[0,r]$ is even closer to $C(r)$ and hence we have an upper bound $\beta(r) < \frac{1}{2r}$ for sufficiently big $r$.

I hope I'm not making some stupid mistake. Suppose $r$ is integral. The distance from the point $[1,r]$ to $C(r)$ is $\sqrt(1+r^2)-r$ which goes to zero as $r$ goes to infinity. For nonintegral $r$ the point $[0,\lceil r \rceil]$ is even closer to $C(r)$ and hence we have an upper bound $\beta(r) < \frac{1}{2r}$ for sufficiently big $r$.