Cornered Again! inspired by "We've got you cornered"

Let the side of the square be $a$ , the radius of first and second two circles mentioned by $r_1$ and $r_2$ respectively.

Then we notice that for the diagonal the first two circles are on:

$\quad \sqrt{2}a = 2(r_1+\sqrt{2}r_1)\quad \Rightarrow r_1 = \dfrac {\sqrt{2}a} {2(1+\sqrt{2})} = (1-\frac {1} {\sqrt{2}})a$

On the other diagonal, half of it is given by:

$\quad \sqrt{(r_1+r_2)^2-r_1^2}+\sqrt{2}r_2 = \dfrac {\sqrt{2}a}{2}$

$\Rightarrow \sqrt{(r_1+r_2)^2-r_1^2} = \dfrac {a}{\sqrt{2}} -\sqrt{2}r_2$

$\Rightarrow r_1^2 +2r_1r_2 + r_2^2 -r_1^2 = \dfrac {a^2}{2} - 2ar_2 + 2r_2^2$

$\Rightarrow r_2^2 -2(a+r_1)r_2 + \dfrac {a^2}{2} = 0\quad \Rightarrow r_2^2 -2(1+1-\frac {1}{\sqrt{2}})ar_2 + \dfrac {a^2}{2} = 0$

$\Rightarrow r_2^2 -2(2-\frac {1}{\sqrt{2}})ar_2 + \dfrac {a^2}{2} = 0\quad \Rightarrow r_2^2 -(4-\sqrt{2})ar_2 + \dfrac {a^2}{2} = 0$

$\Rightarrow r_2 = (2 - \frac {1}{\sqrt{2}} - \sqrt{4-2\sqrt{2}})a$

$\Rightarrow S = \dfrac {2\pi(r_1^2+r_2^2)}{a^2} = \dfrac {2pi(0.2929^2 + 0.2105^2 )a^2}{a^2} = \boxed {0.8174}$

EP in small sketch should be $\dfrac{R}{\sqrt2}$

ABCD is the square, AD the top side. F the center of the First blue circle. FG its vertical radius = R, FE its horizontal radius. K the center of the last green circle. KH its vertical radius = r..... O is the intersection point of the diagonals and also the point of contact of the first two circles. Horizontal OP meets DC at P. Draw KL parallel to AD to meet FG at L. Since it is the ratio, no generality is lost by assuming that the area of the square is 1 unit. So the side of the square =1 unit.
$DE=R.~ EP=\dfrac{FO}{\sqrt2}=\dfrac{R}{\sqrt2} ~~~~..\therefore~\dfrac{1}{2}= DP=R+\dfrac{R}{\sqrt2}\\ \implies~R=\dfrac{1}{2+\sqrt2}....(1) \\LF=FG-LF=R-r...KL=HG=1-R-r...FK=R+r.\\ In~ rt\angle~\triangle~KLF,~~~FK^2=KL^2+LF^2.\\\implies~(R+r)^2=(1-R-r)^2+(R-r)^2\\\implies~r^2 -2r(1+R)+(R-1)^2=0...~~~Solving~ the ~quadratic\\with~+sign,...r=1+R-2\sqrt{R}.........(2)\\ From~ (1) ~and ~(2)~~~Circle ~area~=2*\pi *(R^2+r^2) =0.817. \\\text{Since the square area is 1, this is the ratio.}$