In a square with unit side lengths there are two identical circles that are tangent to each other and two faces of the square. Find the radius? 
In a square with side lengths $1$ there are two identical circles that are tangent to each other and two faces of the square. What is the radius of the circles?

I've spent a bit of time trying to solve this, as part of a larger problem. 

 A: Look at the circle in the upper left-hand-corner.  Let $r$ be the radius of the circle.  If you draw radii to the two sides of the square, you end up with a square of side length $r$.  Moreover, the diagonal of this square is length $r\sqrt{2}$.
Therefore, the diagonal of the large square is composed of two radii and two segments of length $r\sqrt{2}$.  Therefore the diagonal of the square is 
$$
2r(1+\sqrt{2}).
$$
However, the diagonal is also of length $\sqrt{2}$, from here, you can solve for $r$.
A: Imagine the upper left quadrant of the square. Draw horizontal and vertical line segments through the center of the circle between the left/right and top/bottom sides of the quadrant. This creates 4 rectangles in the quadrant (upper left and lower right rectangles are squares). Let $r$ be radius of circle. The area of the four rectangles are: $r^2, \frac{\sqrt{2}}2r^2, \frac{r^2}2, \frac{\sqrt{2}}2r^2$ (why does lower right rectangle have sides of $\frac{\sqrt{2}}2r$?) The total area of quadrant is sum of all 4 rectangle. Area of quadrant is also $\frac14$, so $\frac14 = r^2 + 2 \frac{\sqrt{2}}2r^2+ \frac{r^2}2$ from which $r$ can be calculated.
A: Let $r$ be the radius of our circles.
Thus,
$$r\sqrt2+2r+r\sqrt2=\sqrt2$$ or
$$\sqrt2(1+\sqrt2)r=1$$ or
$$\sqrt2r=\sqrt2-1$$ or
$$r=1-\frac{1}{\sqrt2}.$$
A: Both the diagonals through the square will either touch or dissect at the point of tangent of the circles.  Let the length of the square's side be L.  It follows that the diagonals have length Sqrt(2) x L.  The tangential point is at the mid point of that and Sqrt(2)/2 x L distant from any corner.  This is also the diameter of the circles.
As required the radii of the circles are again half the diameters;
Sqrt(2)/2/2 x L = Sqrt(2)/4 x L.
References; Pythagoras.
Legend;  Sqrt(2) = Square root of 2.  MathJax requires me to Leave the Page.
