Tight Packing Shapes

Let $\angle WDC = \theta$ , then $\tan \theta = \frac {16}8 = 2$ . Let $O$ be the centre and $r$ be the radius of the circle. Then we note that $\angle ODC = \frac \theta 2$ . Then we have $\tan \frac \theta 2 = \frac r{16-r}$ . Using a calculator, we can find $\theta = 1.107148718 \text{ rad}$ , then $\tan \frac \theta 2 = 0.618033989$ and $r = 6.11145618 \approx \boxed{6.11}$ .

Using hand calculations is as follows:

$\begin{aligned} \tan \theta & = 2 \\ \frac {2 \color{#3D99F6}{\tan \frac \theta 2}}{1 - \color{#3D99F6}{\tan^2 \frac \theta 2}} & = 2 & \small \color{#3D99F6}{\text{Note that } \tan \frac \theta 2 = \frac r{16-r}} \\ \frac {2 \cdot \color{#3D99F6}{\frac r{16-r}}}{1 - \color{#3D99F6}{\frac {r^2}{(16-r)^2}}} & = 2 \\ \frac {16r-r^2}{256-32r+r^2-r^2} & = 1 \\ 16r - r^2 & = 256 - 32r \\ r^2 - 48r + 256 & = 0 \\ \implies r & = 24 \pm 8\sqrt 5 & \small \color{#3D99F6}{\text{As }24 + 8\sqrt 5 > 16 \text{, it is rejected.}} \\ & = 24 - 8\sqrt 5 \\ & \approx \boxed{6.11} \end{aligned}$

Using trigonometry.

$\tan \theta=\dfrac{8}{16}$ $\implies$ $\theta=\tan^{-1}\left(\dfrac{8}{16}\right) \approx26.565$

$\alpha=\dfrac{90-26.565}{2} \approx 31.7175$

Then,

$\tan 31.7175=\dfrac{r}{16-r}$

$9.889-0.618=r$

$1.618r=9.889$

$r \approx 6.11$