Placing a circle in a square lattice

Question

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.

Answer 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:

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

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

Answer 2

For the first question it is enough to consider placing the circles in the fundamental domain of the lattice, that is the unit square. The maximum distance to the closest vertex is indeed at the center, so the circle can be placed without containing any of the vertices if and only if $r < \frac{\sqrt{2}}{2}$.

In the second question we would need to first specify what it means to place a random circle of radius $r$ in the plane. I think in the present question the most natural interpretation is again to consider $\mathbb{R}^2$ modulo $\mathbb{Z}^2$, and place the circle uniformly in the fundamental domain of the lattice, i.e. the unit square. If $r < \frac{1}{2}$, then the answer is easy: Each vertex has a quarter circle of radius $r$ of points that have the property that a circled placed at them will contain the vertex. Thus the probability is $\frac{1}{2} \tau r^2$ ($\tau = 2\pi$). If $r \ge \frac{\sqrt{2}}{2}$, the answer is again easy: Any circle will contain one of the vertices. Finally we have the case $\frac{1}{2} \le r < \frac{\sqrt{2}}{2}$. Then it is possible that the same circle contains $2$ vertices, so we have to subtract $4$ times the area of the lens shaped figure between two quarter circles. If I calculated this correctly, the probability should be $$\frac{1}{2} \tau r^2 - 4 r^2 \arcsin\left(\frac{\sqrt{4r^2 - 1}}{2r}\right) + \sqrt{4r^2 - 1}$$