Diagonal Circles in a square

Let the radius of the large congruent circle be $r$ . Considering the diagonal for top-left vertex to bottom-right vertex of the square, we note that:

$2r + 2\sqrt 2 r = \sqrt 2 \implies r = \frac 1{2+\sqrt 2} = 1 - \frac 1{\sqrt 2}$

Now consider the other diagonal:

$\begin{aligned} 2\sqrt 2 R + 2 \sqrt{(r+R)^2-r^2} & = \sqrt 2 \\ 2 R + \sqrt {2(2rR+R^2)} & = 1 \\ \sqrt {4rR+2R^2} & = 1 - 2R & \small \blue{\text{Squaring both sides}} \\ 4rR+2R^2 & = 4R^2 - 4R + 1 \\ 2R^2 - 4(1+\blue r)R + 1 & = 0 & \small \blue{\text{Note that }r = 1 - \frac 1{\sqrt 2}} \\ 2R^2 - (8-2\sqrt 2)R + 1 & = 0 \\ \implies R & = 2 - \frac 1{\sqrt 2} \pm \sqrt{4-2\sqrt 2} & \small \blue{\text{Since }R < 1} \\ & \approx 0.21050 \\ \lfloor 10^4R \rfloor & = \boxed{2105} \end{aligned}$

Draw a diagonal to the square that is tangent to the black circle. By the Pythagorean Theorem, the length of this diagonal is Sqrt(2) (the sides of the square have length 1). Then, the radius of the black circle is the inradius of a triangle with side lengths 1, 1, and sqrt(2). Therefore, if r is the radius of the black circle, we find:

r = 1/(2+sqrt(2))

Let R be the radius of the red circle. Let x be the horizontal distance between the centers of the red and black triangles. Again applying the Pythagorean Theorem, we find:

x^2 + (r-R)^2 = (R+r)^2, or:

x = 2 sqrt(R) sqrt(r).

However, we also see that r+x+R = 1. Therefore, by substitution:

R + 2 sqrt(R) sqrt(r) + r = 1 or

R + 2 Sqrt(R) Sqrt(r) + (r-1)= 0

This last equation is quadratic in Sqrt(R). We can now use the quadratic formula to get R in terms of r (which is known by the argument above). Performing the calculation yields:

R = .210501 (approximately)