I am trying to compute $\pi_1(X)$ where $X$ is the following collection of circles:
$$ X = \left[\bigcup_{(m,n) \in \mathbb{Z}^2} (m+ in) + S^1 \right] \cup \left[\bigcup_{(m,n) \in \mathbb{Z}} \big((m + \tfrac{1}{2}) + i (n + \tfrac{1}{2}) \big) + (1 - \sqrt{2})S^1 \right]$$
I have computed the formula to make the circle tangent patterns to be what it looks like. This is possibly related to the truncated square tiling.
There is an action $\mathbb{Z}^2 \times X \to X$ leaving this space invariant, so this is a covering space $X \to Y$. So perhaps it's sufficient to compute $\pi_1(X/\mathbb{Z}^2)$.
This space is just a graph (consider the intersection points of the circles as vertices and the arcs connecting them as edges). Modding out a spanning tree, it is thus homotopy equivalent to a wedge of circles. It is clear there must be countably infinitely many circles, so $\pi_1(X)$ is a free group on countably infinitely many generators.