Let $r$ be the radius of the smallest circle. Let $O_1$ be the center of the circle of radius 1, $O_2$ be the center of the circle of radius 2, $O_3$ be the center of the circle with radius 3, and $O_r$ be the center of the circle of radius $r$.
If you draw $\triangle O_1O_2O_r$, you can easily verify that it has side lengths 3, $1+r$, and $2+r$. In addition, you can also verify that the cevian $O_rO_3$ to the side with length 3 has length $3-r$. We now have enough information to solve for $r$. If we let $\theta=\angle O_1O_3O_r$, then by the Law of Cosines:
\begin{align}
4+(3-r)^2+4(3-r)\cos\theta &= (1+r)^2\\
1+(3-r)^2-2(3-r)\cos\theta &= (2+r)^2.
\end{align}
Adding the first equation to twice the second yields the equation $6+3(3-r)^2=(1+r)^2+2(2+r)^2$, which has the solution $r=\frac 67$.
Alternatively, you can use Descartes' Theorem, in particular, the following formula:
$$\frac1r=\frac 11+\frac12-\frac13\pm2\sqrt{\frac1{1\cdot2}-\frac1{2\cdot3}-\frac1{3\cdot1}}=\frac 76,$$
so $r=\frac67$.
rather than cranking through the algebra it's probably better to solve this as a "secular" geometry problem and then translate back to the plane.
let $d=|\textbf{x}_1-\textbf{x}_2|$
Define $\alpha$ as the angle measured from the line of centres of the radius to the first point of tangency ( where the tangent lies above both circles )
Define $\beta$ as the angle measured from the line of centres of the radius to the second point of tangency ( where the tangent lies above above the left circle and below the right circle. )
you get $ \cos \alpha = \frac{d}{r_1 - r_2}$ and $ \cos \beta = \frac{r_1 + r_2}{d}$
so the four points of tangency on the left circle have angles of $\pm \alpha$ and $\pm \beta$ relative to the line of centres. The corresponding points on the right circle are at angles $ \pm (\pi - \alpha) $ and at $\mp \beta$, again relative to the line of centres.
to find the co-ordinates of the first point of tangency the only extra bit of info you need is the angle of the line of centres to the horizontal.
$\tan \theta = $ the slope of the line joining $\textbf{x}_1$ and $\textbf{x}_2$
$ \textbf{t}_1= \textbf{x}_1 + r_1(cos( \alpha + \theta), sin(\alpha + \theta))$
follow an analogous procedure to determine the other seven points.
Best Answer
Adjusting Matthew's figure:
$\hspace{2cm}$
Use Cosine theorem: $$\frac{FC}{DC}=\frac{\sqrt{3^2+3^2-2\cdot 3^2\cdot \cos \beta}}{\sqrt{2^2+2^2-2\cdot 2^2\cdot \cos (180^\circ-\beta)}}=\sqrt{\frac{18-18\cdot \frac15}{8+8\cdot \frac15}}=\sqrt{\frac32}.$$ Note: $\cos \beta =\frac{BE}{AB}=\frac{FB-FE}{AC+BC}=\frac{FE-AD}{AC+BC}=\frac{3-2}{2+3}=\frac15$.