Find the radius of the smallest circle in the figure

Let $O$ be the center of the orange circle. Then, since the side length of the square is $2(6 + 3) = 18$ ,

$\tan(\angle OCS) = \dfrac{6}{9} = \dfrac{2}{3} \Longrightarrow \angle OCS = \tan^{-1}\left(\dfrac{2}{3}\right)$ .

Now as $PC$ is tangent to the orange circle we see that $\angle PCO = \angle OCS = \theta$ , and thus

$\angle PCN = \dfrac{\pi}{2} - 2\theta \Longrightarrow \tan(\angle PCN) = \cot(2\theta) = \dfrac{1}{\tan(2\theta)}$ .

Now $\tan(2\theta) = \dfrac{2\tan(\theta)}{1 - \tan^{2}(\theta)} = \dfrac{2 \times \dfrac{2}{3}}{1 - \left(\dfrac{2}{3}\right)^{2}} = \dfrac{12}{5}$ .

Next, $|PN| = 12\tan(\angle PCN) = 12 \times \dfrac{1}{\dfrac{12}{5}} = 5$ ,

so $\Delta PCN$ is a $5/12/13$ right triangle, and thus the inradius is $\dfrac{5 \times 12}{5 + 12 + 13} = \boxed{2}$ .