Three circles within a larger circle

The middle stanza of Soddy's Kiss Precise gives the formula:

Four circles to the kissing come.
The smaller are the benter.
The bend is just the inverse of
The distance from the center.
Though their intrigue left Euclid dumb
There's now no need for rule of thumb.
Since zero bend's a dead straight line
And concave bends have minus sign,
The sum of the squares of all four bends
Is half the square of their sum.

Applied here it says $$\frac 3{r^2}+\frac 1{R^2}=\frac 12 \left(\frac 3r-\frac 1R\right)^2\\\frac 3{r^2}+\frac 1{R^2}=\frac {9}{2r^2}+\frac 1{2R^2}-\frac 3{rR}$$ As all the we can get is the ratio, let $r=1$ and we have $$3+\frac 1{R^2}=\frac 92+\frac 1{2R^2}-\frac 3R\\0=3R^2-6R-1\\R=\frac 16(6\pm\sqrt{48})=\frac 13(3\pm2\sqrt{3})$$ and we want the plus sign.

Call the radius of the smaller circles $r$. Their centers form an equilateral triangle of side $2 r$. The centers of the small circles are at a distance of $R - r$ from the large circle's center, and $r$ from the large circumference. The tangency points of the smaller and large circle are also an equilateral triangle. I believe that drawing all the triangles mentioned gives you enough in terms of angles to find relations among $r$, $R$ and $R - r$ to get $r$ by trigonometry.

Perhaps you can use Descartes' Theorem here: http://en.wikipedia.org/wiki/Descartes%27_theorem

Note: There is also a circle internally tangent to the three tangent circles.

(I get (-3 + 2 √3)R = r for the relationship between the outer circle of radius R to the inner circle(s) of radius r.)