Tangent circles in geometric progression.?

I think you need to clarify whether the circles in geometric progression are tangent to each other or not. Makes a big difference.

Let's assume that they don't have to be. Then 3 variables describe the circles in geometric progression, namely r, and x, y in z = x + iy. For 3 circles, there will always be a circle(s) tangent to all three, defined by R, X, Y. Essentially, we could have a set of 6 equations to solve for 6 unknowns, r, x, y, R, X, Y, for a "maximum length of 6" but it is very likely that the few solutions that exist are going to be trivial cases. So, I'd put my money on 5 circles being the maximum, although I only have an hazy idea of how to start cranking out some numbers, and I will need some time. Would you accept an approximation, as "exact solutions" seems hopelessly out of reach?

It is possible that there is some kind of a Steiner chain situation here where you can have as many as you want, but I kind of doubt it.

Edit: This mention of "loxodromic sequence" of circles probably doesn't count because the centers aren't of the form z^n, if I understand the problem correctly.

Edi 2: No, the only thing I know for sure is a sequence of 4 circles. I'm sure 5 is possible. 6 seems like a stretch, except in trivial cases, such as r = 1 being forced or that all of the circles pass through (0,0). Anything more than 6 will require a whole rethink of this problem. I am assuming that the centers of the circles can be mapped on the complex plane as z^n, instead of just anywhere. Also, no, the Steiner chain itself cannot ever be a solution, I was merely referring to the possibility of something like it, although I have no idea of how it would even work. I'm pretty sure the number of circles is limited, and my best guess right now is just 5.

Edit 3: See link for graphic of a sequence of 5 circles, where

z = 1.27 Cos(220) + 1.27 Sin(220) i

r = 0.87

R = 2.444

X = 0.425

Y = -0.092

This is very loosey-goosey stuff, I'm not even sure if there's an exact solution in this arrangement.

Edit 4: gianlino, where I am right now, it's 3 am in the morning. See next graphic, a nicer, more expected arrangement, but no 6. I have no idea how to bring in the 6th one. Tomorrow's Sunday.

Edit 5: 2nd pic specs:

z = 1.66 Cos(121.7) + 1.66 Sin(121.7) i

r = 1.455

R = 5.941

X = 2.483

Y = -1.565

(fixed specs on R circle)

Edit 5: 3rd pic specs:

z = 1.40 Cos(128.5) + 1.40 Sin(128.5) i

r = 1.10

R = 3.398

X = 0.989

Y = -0.253

It isn't too much trouble finding solutions where there are 5 circles tangent to a common one, maybe I'll pick one and perfect it to a higher degree of accuracy. However, getting all 6 seems hopeless, and more than 6 seems impossible. I can't visualize how it would even look.

Edit 6: Yes, it is possible, for each angle (argument), to iteratively work out a solution where the first 5 are tangent to a common circle, leaving the 6th one "hanging out". Then it's just a matter of checking out all the possible angles from 0 to 180. If the 6th one won't come in, then there is no solution that I can see. At least for the case where 3 circles are inside and 3 circles are outside the common circle tangent to all. Now, let me try a whole different approach.

Edit 7: All right, have a look at the 4th solution, 4th link:

z = 0.725 Cos(142) + 0.725 Sin(142) i

r = 0.545

R = 0.213

X = 0.036

Y = -0.0078

It looks like a 6 circle solution, but I don't believe it's a real solution. If it is, then there is something deeper going on that I don't understand yet.

Edit 8: Let's see if I can reverse engineer the pic you've posted. This is the kind special non-trivial case where we have exactly 6 circles sharing a common tangent circle, in an infinite sequence of sequences. The question is, are the centers of those circles of the form z^n? I'll find out, maybe later tonight or tomorrow.

The point is, there's an infinity of solutions where there are 5 circles, but it was a question whether there exists any special 6 circle solutions that aren't trivial. Now the question is, are there any more? There can only be a finite number of such solutions. I must admit this is very fascinating.

Edit 9: In order for to have self-similar sequences of sequences of circles, as depicted by your picture, the radii of the circles need an extra variable, namely radius = a r^n, not just r^n. That means 7 degrees of freedom, not 6. So, there should be an infinity of solutions in that case. Yes, that is the classic definition of a geometric progression, but I took R_n =r^n to mean that a = 1. Anyway, I think I am running out of space here, Y!A is going to truncate me soon.

Coxter's Loxodromic sequence of tangent circles Is an infinite sequence of circles arranged such that any 4 circles in the sequence are pairwise mutually tangent. That means that each circle in the sequence is tangent to the 3 circles that precede it and also to the three circles that follow it. Hence the maximum length of a geometric sequence of circles such that there exists a circle tangent to all of them and not included in sequence is 6.