Prove: at most two circles are needed to be tangent to all the circle sequence

Question

Construct a circle sequence $\{C_n\}$ (e.g., blue in the figure below) in 2D Cartesian coordiante system as:

1. the $x$-coordiantes of centers of all the circle $C_n$ are $\frac{1}n$;
2. all the circles $\{C_n\}$ are above and tangent to $x$ axis;
3. all the circles $C_n$ are tangent to their neighbour circles, i.e., $C_{k}$ is tangent to both $C_{k-1}$ and $C_{k+1}$, $\forall k>1$

Prove: there needs at most two circles (e.g., the red cirlces in the figure below) to be tangent to all the circles $\{C_n\}$.

update:

I mean there exists a set of circles such that no circle in the defined sequence is not tangent to a member of the circle set. Now prove the least upper bound of the set number is 2

I was considering to use inversion to transform the two red circles into two horizontal lines, but failed to obtain an explicit form

A comment from mathoverflow which I have difficulty in understanding:

This requires almost nothing about the set of circles Cn except its infinitude. The only other fact we need is that they're not all tangent to each other at the same point. Indeed for any three circles Σ1,Σ2,Σ3 there are finitely many circles C tangent to each Σi (i=1,2,3), unless the Σi are tangent to each other at the same point P and then every other C must also be tangent to them at P. [In fact in the absence of such P there are at most eight C's, so as soon as we reach n=9 we're already done.] – Noam D. Elkies Jun 8 at 2:44

Answer 1

The map $f(z)=\frac 1{\overline{z}}=\frac z{|z|^2}$ works. For example, it’s easy to check that the circle $|z-ri|=r$ of radius $r>0$ is mapped to the horizontal line $y=\frac 1{2r}.$

Now $f$ retains the conformal behavior (up to reflection) and maps circles (or lines) to circles (or lines). The points $\frac 1 n,n\in{\mathbb N}$ are mapped to $n,n\in{\mathbb N}.$ To check the claim, it suffices to examine those circles tangent at $x=n$. For simplicity, one looks at circles tangent at $x=1$ of radius $r_1$ and at $x=2$ of radius $r_2$ such that they are mutually tangent. By Pythagorean theorem, $r_1,r_2$ satisfy the relation $$1+(r_2-r_1)^2=(r_1+r_2)^2,$$ hence $$r_1r_2=\frac 14.$$ It follows that $r_2r_3=\frac 14,$ etc. Consequently $$r_1=r_3=r_5=\cdots, r_2=r_4=r_6=\cdots,$$ hence they are at most all tangent to two horizontal lines. Applying the inverse transformation then proves the claim. QED

Remark. In the particular case when $r_1=\frac 12,$ the circles will all be tangent to one horizontal line with equation $y= 1.$ Applying inverse transformation, this corresponds to the circle tangent at $x=1$ with equation $(x-1)^2+(y-1/2)^2=1/4$ and the circle tangent at $0$ with equation $\left|z-\frac 12i\right|=\frac 12$ that is tangent to the whole family of circles.

Answer 2

This is what I had in mind. If you invert your original picture in the unit circle around the origin, the $x$ axis is mapped to itself, all lines and circles are mapped to lines or circles, and all tangency is preserved. So you get circles tangent to the $x$ axis at positive integers.

The circles might all have diameter $1,$ that is the top row in purple ink.

The circles might have varying diameters. However, there is then an explicit relationship of diameters at consecutive integers, as that pair of circles must be tangent. What is the relationship of diameters in my picture?

The extreme is in the bottom row in red.

In these rows, there are evident horizontal lines that are tangent to all these circles, I did not draw those. Those lines are mapped back to something nice when inverting my picture back to your original setting

Below is the middle row of circles, done over to include the new horizontal lines that are, together, tangent to all the circles