More Circles and Squares !

Let $2z$ be the length of a side of square base $ABCD$ and let $A:(0,0), B:(0,2z),$

$C:(2z,2z), D:(2z,0)$ and $E:(z,2z)$ and $O:(x,y)$ .

$(1): (x - z)^2 + (y - 2z)^2 = r^2$

$(2): x^2 + y^2 = r^2$

$(3): (x - 2z)^2 + y^2 = r^2$

Subtracting $(2)$ from $(1) \implies 2zx + 4zy = 5z^2$

and

Subtracting $(2)$ from $(3) \implies 4z(z - x) = 0$ and $z \neq 0 \implies x =z \implies$

$y = \dfrac{3}{4}z \implies r = \dfrac{5}{4}z \implies$ the equation of the circle is

$(x - z)^2 + (y - \dfrac{3}{4}z)^2 = \dfrac{25}{16}z^2 \implies$

$y = \sqrt{\dfrac{25}{16}z^2 - (x - z)^2} + \dfrac{3}{4}z$ which is the portion of the circle needed to obtain the above area.

The area $I_{1} = 2\displaystyle\int_{0}^{z} 2z - ( \sqrt{\dfrac{25}{16}z^2 - (x - z)^2} + \dfrac{3}{4}z) \:\ dz =$

$2\displaystyle\int_{0}^{z} \dfrac{5}{4}z -\sqrt{\dfrac{25}{16}z^2 - (x - z)^2} \:\ dz$

Let $x - z = \dfrac{5}{4}z\sin(\theta) \implies dx = \dfrac{5}{4}z\cos(\theta) d\theta \implies$

$I_{1} = 2(\dfrac{5}{4}z^2 - \dfrac{25z^2}{32}(\arcsin(\dfrac{4(x - z)}{5z}) + \dfrac{4(x - z)\sqrt{25z^2 - 16(x - z)^2}}{25z^2})|_{0}^{z})$

$= 2(\dfrac{5}{4} - \dfrac{25}{32}\arcsin(\dfrac{4}{5}) - \dfrac{3}{8})z^2 =$ $(\dfrac{5}{2} - \dfrac{25}{16}\arcsin(\dfrac{4}{5}) - \dfrac{3}{4})z^2$

$I_{2} = 2\displaystyle\int_{0}^{z} \sqrt{\dfrac{25}{16}z^2 - (x - z)^2} - \dfrac{3}{4}z \:\ dx$

Letting $x - z = \dfrac{5}{4}z\sin(\theta) \implies dx = \dfrac{5}{4}z\cos(\theta) d\theta \implies$

$I_{2} = (\dfrac{25}{16}\arcsin(\dfrac{4}{5}) - \dfrac{3}{4})z^2$

$\implies$ the total area $A = I_{1} + I_{2} = z^2 \implies \dfrac{A_{\square{ABCD}}}{A} = \boxed{4}$