Determine relative position of 3 large (equal) circles and 1 smaller circle within a minimum enclosing circle I want to visualize the position of $3$ large equal circles with radius $r_1$ and $1$ smaller circle with radius $r_2$. The circles represent three wire conductor phases and one smaller neutral conductor. The enclosing circle represents the minimum wire bundle diameter.
I have tried solving the problem using trigonometric functions and algebra but so far unfortunately with no success. I would like to be able to determine the position of all $4$ circles from the enclosing circle center. This means that I am interested in dimensions: W, X, Y and Z. I am expecting that all dimensions can be written as a function of $r_1$ and $r_2$. In the picture below I am interested mainly in cases 1 and 2, case 3 is optional.
Thanks in advance!

 A: For Case 1, using Descartes' Kissing Circles Theorem, we can determine the maximum possible value of $r_2$.  From the $r_1$ circles forming an equilateral triangle, it follows that the enclosing circle has a radius of
$ R = r_1 + \dfrac{2}{\sqrt{3}} r_1 $
If the small circle (of radius $r_2$) is tangent to the left and right $r_1$ circles then
$ \dfrac{1}{r_2} = \dfrac{2}{r_1} - \dfrac{1}{(1 + \dfrac{2}{\sqrt{3}}) r_1 } + \dfrac{2}{r_1} \sqrt{ 1 - \dfrac{2}{1 + \dfrac{2}{\sqrt{3}} } } $
This gives:
$ r_2 = 0.482672825 \ r_1 $
If $r_2$ is less than or equal to the above value, then we have Case 1.  If $r_2$ is greater than the above value, then we have Cases 2 and 3.
For Cases 2 and 3, and using the variables used in the supplied figure in the OP, we have the following equations
The centers of the four circles are
Top:  $(0, Z)$
Right: $(W, Y)$
Left: $(-W, Y)$
Bottom: $(0, X)$
And we have an additional variable which is $R$ the radius of the encompassing circle.  So a total of $5$ variables.  Now we can write the following $5$ equatons:
$ R = r_1 + Z  \hspace{30pt} (1) $
$ R = r_1 + \sqrt{ W^2 + Y^2 }\hspace{30pt} (2) $
$ 4 r_1^2 = W^2 + (Z - Y)^2 \hspace{30pt} (3) $
$ (r_1 + r_2)^2 = W^2 + (Y - X)^2 \hspace{30pt} (4)$
$ R = r_2 - X \hspace{30pt} (5)$
These equations can be solved numerically without much effort using the Newton-Raphson multivariate method, and it converges to the solution with a reasonable initial guess of the solution, in less than 10 iterations.  This is a sample result of what I got with my implementation:

