Squares at least partially inside

Toss a circle of whatever radius you like onto a square grid. Color the squares that are fully inside the circle, and the squares touching or intersecting the boundary of the circle. Call the colored shape an "at least partially covered" shape. Toss another circle on the grid, but this time, only color the squares that are fully inside the circle. Call this a "fully inside" shape.

What is the largest at least partially covered square you can make? To answer this question, we can ask a related question:

Draw a large corner. In the corner, wedge a unit square. What is the radius of the circle that touches both walls of the corner, and simultaneously, the protruding vertex of the square?

Draw such a circle that touches the walls and protruding vertex. This circle needn't be exact. Draw a dashed line from the center of the circle to the corner's vertex. This dashed line is the diagonal of a square having radius r, thus, the length of the diagonal is \(r\sqrt{2}.\) If we remove \(r\) from this length, we're left with the diagonal length of the unit square. But this also has length \(\sqrt{2}.\) Thus, the radius of the circle we're after is \(r\sqrt{2} - r = \sqrt{2}.\) Doing a little bit of algebra yields

$$\begin{align} & r\sqrt{2} - r = \sqrt{2} \\[0.75em] & r(\sqrt{2} - 1) = \sqrt{2} \\[0.75em] & r = \dfrac{\sqrt{2}}{\sqrt{2} - 1} \approx 3.41 \end{align}$$

This means the largest at least partially covered square must be less than \(2 \cdot 3.5 = 7\) wide, and so the largest square you can make is \(6 \times 6.\)