Placing a circle in a square lattice Two part question. Consider the square lattice $\mathbb{Z}^2$:

Imagine you are going to place a circle of radius $r$ somewhere in $\mathbb{R}^2$.
Question 1:
What is the radius of the largest circle that cannot be placed anywhere in $\mathbb{R}^2$ without overlapping or containing any of the points in $\mathbb{Z}^2$?

 I'm fairly certain the answer is $r < \frac{\sqrt{2}}{2}$. If you place a circle with center, for example $\left(\frac{1}{2}, \frac{1}{2}\right)$, computing the radius that would touch the four corners is quite easy, but I don't know how to prove that this is the best place to do it, as intuitively obvious as it might be.

Question 2:
A circle with radius $r$, such that $0 < r \leq \frac{\sqrt{2}}{2}$ is randomly placed somewhere in $\mathbb{R}^2$. What is the probability, as a function of $r$, that the circle contains or overlaps at least one point in the lattice?

 My suspicion is to try to only consider a limited subset of the 2D plane, e.g. $-1 < x < 1$ and $-1 < y < 1$, and then do something with ratios of the area of the circle and the area of the limited region, but I'm not exactly sure what I would do with that. Obviously the probability is 1 for all $r \geq \frac{\sqrt{2}}{2}$, because of question 1.


You can consider the nearest point in the lattice to the circle center. This induces a Voronoi partition of the plane, which, in this cases, corresponds to squares.
This already answers question 1: the biggest distance to the nearest lattice point is half the diagonal of the unit square.
And it already answers question 2: considering that the probability in uniform over this square, the probability that it does not touch the nearest point (and hence, any point) is given by:


*

*If $r \le1/2$: area of the square substracted from a circle: $p(r)=1-\pi r^2$

*If $r > 1/2$: area of the four corners
