Arabian circles This lovely pattern comes from a show about arabian patterns.  Thus the title.

It includes circles touching 8, 7, 6, 5 and 4 other circles.
The question: what are the exact sizes of the circles in decreasing order?
I know the numerical solution, I am interested in the exact expressions.  I am not sure whether such an expression exists.
For the discussion purpose, lets call the radii a, b, c, d and e for the circles with resp. 8, 7, 6, 5 and 4 neighbors.  Let's assume the period of the pattern is 2 units.
There is no catch in the picture.  Where 2 circles seem to touch, they do touch.  And you can assume perfect symmetry horizontal, vertical and diagonal.
PS: sorry if it is too easy for this forum, I believe it is too difficult and too mathematical for the puzzling forum.
 A: "Just" solve this system:

$$\begin{align}
|BC| &= b+c \\
|BD_1| &= b+d \\
|CD_2| &= c+d \\
|D_1 D_2| &= 2 d \\
|D_2 E| &= d + e
\end{align}$$
where, say, 
$$A = (0,0) \qquad B = (a+b,0) \qquad C = \frac{(a + c )\sqrt{2}}{2} (1,1)$$
$$D_1 = (a+2b,x) \qquad D_2 = (x,a+2b) \qquad E = (a+2b)(1,1)$$

Edit. Numerically, taking $a=1$, we have 
$$\begin{align}
b &= 0.7037139\dots \\
c &= 0.5493113\dots \\ 
d &= 0.7792658\dots \\ 
e &= 0.3227824\dots \\ 
x &= 1.3053795\dots
\end{align}
$$
which, according to Mathematica, is the only solution in positive real values.
Symbolically, well ... There's an unattractive degree-6 polynomial involved, and I don't really want to type it in right now. I'll revisit this later.
A: Do we know that the pattern really fits like it looks?  Two starts, but too long to comment:  Looking at the smallest circle and two of its neighbors, you have a $90-45$ triangle joining the centers with legs $b+e$ and hypotenuse $2b$.  This gives $e=(\sqrt 2-1)b$  Looking at the largest circle, you have a triangle with sides $a+c,a+d,c+d$ with the angle opposite $c+d$ being $45^\circ$.  This gives $(c+d)^2=(a+c)^2+(a+d)^2-2(a+c)(a+d)\frac {\sqrt 2}2$ from the law of cosines.  Unfortunately, the other circles do not have nicely symmetric surrounds, so getting the other two equations will not be as easy.
