We've got you cornered .....

Let $r$ be the radius of each of the two "whole" circles and $R$ be the radius of each of the two quarter-circles. Also, without loss of generality, let the square in question be a unit square. Then the combined areas of the circles and quarter-circles will be the desired ratio $S$ .

Looking first at the diagonal of the square and the two whole circles, we see that the center of each circle is a distance of $\sqrt{2}*r$ from the nearest corner. So as the centers of the circles are a distance $2r$ apart, we have that

$2*\sqrt{2}*r + 2r = \sqrt{2} \Longrightarrow r = \dfrac{2 - \sqrt{2}}{2}.$

Now form a right triangle with vertices being the center of one of the whole circles, the center of one of the quarter-circles and the last being the nearest point of tangency of the whole circle to the hypotenuse. The hypotenuse will then have length $r + R$ and the legs will have lengths $1 - r$ and $r$ . By Pythagoras we then have that

$(r + R)^{2} = (1 - r)^{2} + r^{2} \Longrightarrow R = \sqrt{2 - \sqrt{2}} - \dfrac{2 - \sqrt{2}}{2}.$

The combined areas of the whole circles and quarter-circles, and hence the desired ratio, will thus be

$S = 2\pi*r^{2} + \dfrac{\pi}{2}R^{2} = 0.8896629....$ ,

and so $\lfloor 10000*S \rfloor = \boxed{8896}$ .