Eccentered Circles - determine space between circle at a given location I need to figure out a way of calculating the dimensions x and y as shown on the attached image. I know the angles (in the example the inner circle is broken into 6 - 60 degree angles). I also know the diameters of both circles.
I have tried a bunch of approaches but I think I am missing some simple point that is not letting me see the solution.
David

 A: Let $O_1$ and $O_2$ be the centres of the small and large circles, and let $r_1$ and $r_2$ be their radii. Let $P$ be the point on the large circle in the top right (so $x$ measures the distance from the small circle to $P$).
Now consider the triangle $PO_1O_2$. Its side lengths are $r_2-r_1$, $r_2$, and $r_1+x$. The angle $\angle PO_1O_2$ is known (in this case, $60^\circ$). Now use the Cosine Law to find $x$.
A: choose the units so that the smaller circle has radius $1$ and the bigger one has radius $R.$ let the center of the smaller circle be $O,$ the common point of contact be $A,$ and the point $C$ such that $AC$ is the diameter of the bigger circle. the points $B, D$ are on the bigger circle and $BOD$ is a line. 
applying the cosine rule to the triangle $AOB$ gives $$AB^2 = 1 + (1+x)^2 + (1+x)$$ 
applying the cosine rule to the triangle $COD$ gives $$CD^2 = (2R-1)^2 + (1+y)^2 -(2R-1)(y+1)$$ 
the two triangles $AOB, DOC$ are similar so that $$\dfrac{1}{1+y} =\dfrac{1+x}{2R-1} = \dfrac{BA}{CD}  \tag 1$$
we can solve $(1)$ for $x$ and $y.$
