Using the previous notation, where A6, B6, and A2 are at (0, 0, 6),

(0, 0, -6), and (0, sqrt(28), 0) respectively, and with B2 on top and C being the ball in the middle, we again first get the coordinates of B2. 

Including O, the origin, we have another (6, sqrt(28), 8) right triangle: OA6B2. 

OA6 = 6, OB2 = sqrt(28), and A6B2 = 6+2 = 8, with A6OB2 being the right angle. Dropping a line down from B2 to be perpendicular to the x-y plane, we see it hits the plane at (0, B2(y), 0) which is the right angle corner of yet two more right triangles. The height of the line is B2(z). The point on the plane (0, B2(y), 0) forms a right triangle with hypotenuse OB2 = sqrt(28) and another right triangle with hypotenuse A2B2 = 2+2 = 4.  These give: 

B2(z)^2 + B2(y)^2 = 28, and 

B2(z)^2 + (sqrt(28) - B2(y))^2 = 16

These solve as:

 

B2(y) = 10/sqrt(7) = 3.77964473...   and

B2(z) = 4 sqrt(6/7) = 3.70302804...

Second, we get the center of C. We see we have three unknowns: Cy and Cz (with Cx = 0), and Cr, (we will call them: y, z, and r). We also have three equations from the distances to the known centers of the other 3 balls:

 

r + 6 = sqrt (6^2 + y^2 + z^2)    [CA6]

r + 2 = sqrt (y - 10/sqrt(7) + (z - 4 sqrt(6/7)^2 )  [CB2]

r + 2 = sqrt (y-sqrt(28) + z^2)    [CA2]       

Equating the RHS of the 2nd and 3rd equations gives z = y/sqrt(6)

The 1st and 3rd equations together now have only two unknowns: y and r. With a bit o' Wolfram alpha to save us from tedium: 

r = +/-9 sqrt(2) - 12 

Choosing the positive root,

Cr = 0.72792206...

y also comes out from this:

C2(y) = 18(sqrt(2)-1)/sqrt(7) = 2.8180449...

so:

C2(z) = B2(y)/sqrt(6) = 1.150461725...

It is nice that all these results agree with the previous post where I did it simply by trying numbers and iterating, and also that the error bars previously quoted are here proven OK.