A square with side length $7$ contains four congruent squares with disjoint interiors, each sharing a corner with the big square, together with a circle centered at the middle of the big square. The circle's diameter equals the common side length of the four squares, and the circle passes through the vertex of each square nearest the center, as shown. What is the common side length?

1. Let the common side length be $x$, so the circle has radius $x/2$. The corner square nearest a given corner has its inner vertex at distance $\sqrt2\left(\tfrac72-x\right)$ from the center of the big square (half the big square's diagonal, minus the diagonal offset of the corner square).
2. Since the circle passes through that vertex, $\dfrac{x}2=\sqrt2\left(\dfrac72-x\right)$, so $x+2\sqrt2\,x=7\sqrt2$.
3. Solving, $x=\dfrac{7\sqrt2}{1+2\sqrt2}=\dfrac{7\sqrt2(2\sqrt2-1)}{(2\sqrt2)^2-1^2}=\dfrac{28-7\sqrt2}{7}=\boxed{4-\sqrt2}$.