Just know the diagonal . .

First of all, the point where the two middle circles are tangent is the center of the square. Thus, if we assume the square has a side length of one, the distance from the center of the square to the bottom left vertex is $\frac{\sqrt{2}}{2}$ . Next, draw perpendicular's from the bottom most circle's radius, which I will name $O$ , to its points of tangency with the square. Realizing that these perpendiculars are equal in length because they are radii, the distance from $O$ to the corner of the square must be $\sqrt{2}r$ where $r$ is the radius of the individual circles. Therefore, $\begin{aligned} 3r + \sqrt{2}r &= \frac{\sqrt{2}}{2} \\ r(3+ \sqrt{2}) &= \frac{\sqrt{2}}{2} \\ r & = \frac{3\sqrt{2} - 2}{14} \end{aligned}$ Then, solving for area of all four circles leads to $\begin{aligned} A_{1} &= \bigg(\frac{3\sqrt{2} - 2}{14}\bigg)^{2}\pi \\ A_{1} & = \frac{11-6\sqrt{2}}{98}\pi \\ A_{4} & = \frac{22-12\sqrt{2}}{49}\pi \\ \end{aligned}$ The ratio of the square's area to the area of the four circles is then $\frac{1}{ \frac{22-12\sqrt{2}}{49}\pi} = \frac{11+6\sqrt{2}}{2\pi}$ This implies then that $a + b + c = 11 + 6 + 2 = \boxed{19}$ .