Half-Inscribed

Let the radius of the circle be $r$ . Then the side length of square $COFG$ is $r$ and the area of the yellow region is

$\begin{aligned} r^2 - \frac {\pi r^2}4 & = \frac {36-9\pi}{16} \\ \blue{16r^2} - 4\pi r^2 & = \blue{36} - 9\pi \\ \implies r^2 & = \frac {36}{16} \\ r & = \frac 64 = \frac 32 \end{aligned}$

Now note that $\triangle ABC$ and $\triangle AOD$ are similar. Then we have

$\begin{aligned} \frac {BC}{AC} & = \frac {OD}{AD} \\ \implies BC & = \frac {OD}{AD} \times AC \\ & = \frac {OD}{AD} (AO + OC) \\ & = \frac {OD}{AD} \left(\sqrt{OD^2+AD^2} + OC\right) \\ & = \frac {\frac 32}2 \left(\sqrt{\left(\frac 32\right)^2 + 2^2} + \frac 32\right) \\ & = \frac 34 \left(\frac 52 + \frac 32\right) \\ & = \frac 34 \times 4 = \boxed 3 \end{aligned}$

Let the radius of the circle = $x$

Area of the shaded region $x^2 - \frac{π}{4}x^2 = \frac{36-9π}{16}$ $x= \sqrt{\frac{9(4-π)}{4(4-π)}} = \frac{3}{2}$ Now , Join $OD$ to get a right angle triangle

So , $AD^2 +OD^2 = (AE+EO)^2$ $2^2 + (\frac{3}{2})^2 = (AE+\frac{3}{2})^2$ $AE^2 +3AE-4 =0$ $AE= 1$ In triangle $ABC$ , $BD= BC =y$ ( they are tangent from the same point )

So , $(2+y)^2 = (\frac{3}{2}+\frac{3}{2} + 1)^2 + y^2$ $4+4y= 16$ $y= \boxed{3}$

Refer to Ethan Mandelez's diagrams for this answer.

Let $OC = r$ . Then $r^2 - \frac{1}{4}\pi r^2 = \frac{36-9\pi}{16} \Rightarrow r^2 \left(1 - \frac{1}{4}\pi \right) = \frac{1}{16} (36 - 9 \pi)$ , so $r^2 = \frac{1}{16} \times 36 \Rightarrow r = 1.5$ (since $r$ cannot be negative).

Thus $AD^2 + DO^2 = AO^2$ , but since $DO = r$ as well, $AO = \sqrt{2^2 + 1.5^2} = 2.5$ .

Now here is where our solutions differ. Since $\Delta AOB$ and $\Delta BOC$ have the same height, through the formula $A = \frac{1}{2}bh$ , the ratio of their areas must be the ratio of their bases. Thus:

$\frac{\Delta AOB}{\Delta BOC} = \frac{AO}{OC}$ $\Rightarrow \frac{\frac{1}{2} (2+x)(1.5)}{\frac{1}{2}(1.5)(x)} = \frac{2.5}{1.5} \tag{\text{tangents from the same point are equal}: DB = CB = x}$ $\Rightarrow \frac{3 + 1.5x}{1.5x} = \frac{2.5}{1.5} \tag{\text{cancel and simplify}}$

and the rest is straightforward: $\Rightarrow 1.5(3 + 1.5x) = 2.5(1.5x)$ $\Rightarrow 4.5= 1.5x, x = \boxed{3}.$