Let the center of the circle of radii $100$ have center $(m+r,n+s)$ where $m,n \in \mathbb Z$ and $0 \le r,s \le 1$. By symmetry, it is enough to prove it in the case where $s \ge 0.5$. In that case look to the arc of the circle in the region of $y=n-99$. As $n+s-100+0.5 > n-99>n+s-100$ , it is enough to see the difference $\Delta y \le 0.1$ when $\Delta x=1$ for $2$ points on the arc. This is true as $( \sqrt{99.75}-1)^2+(99.5+0.1)^2 > 100^2$.