Assume that the radius of the bigger circle is $R=1$, we compute first the radius $G$ of the green circle from the picture:
There are two similar triangles with the same marked violet angle, the similarity (or the same sine of this angle) gives:
$$
\frac{R-2G}R=\frac G{2R-G}\ ,\qquad\text{ so }\qquad
\bbox[lightgreen]{\color{green}{\ G=\frac 12(3-\sqrt 5)\ }}\ .
$$
Now we have a special situation to be arranged inside the smaller part of the disk with radius $R=1$ delimited by the pink chord in the picture. We need to fit inside one of the remained empty white space the blue, and the red circle. We use inversive geometry, centered in the point $\Omega$ in the picture below.
In the second part we compute the proportions $G:b:r$ between the sizes of the green, the blue, and the red circle.
The picture had a lot to do with the details, but the inversive argument is simple:
Let $\color{magenta}{\Omega}$ be the point on the pink chord and the violet arc delimiting the region where the green, blue, red circles live in. The other end of the chord is $\color{magenta}{\Upsilon}$.
Let $\color{magenta}{\Psi}$ be the other end of the diameter through $\color{magenta}{\Omega}$. Let $A,B,C$ be the tangency points of the
green, blue, red circles shown in the picture with the line $\color{magenta}{\Omega\Upsilon}$.
The picture shows, but we do not need, the points $S,T,U$ of tangency of the
green, blue, red circles with the big initial violet circle
of radius one.
We are performing an inversion, denoted by star, $$X\to X^*\ $$ with center in $\color{magenta}{\Omega}$ and with power $\color{magenta}{\Omega\Upsilon^2}
=4\;\Omega A^2=4(1^2-(1-2G)^2)=16(\sqrt 5-2)$.
- So $\color{magenta}{\Upsilon}$ is invariated, $\color{magenta}{\Upsilon^*=\Upsilon}$.
- The half line $\Omega ABC\Upsilon$ is mapped into itself, and the points
$A^*,B^*,C^*$ are beyond $\Upsilon$. Since $A$ is the mid point of $\Omega\Upsilon$ the transformed $A^*$ is the reflection of $\Omega$ in $\Upsilon$.
The points $\Omega,A,B,C,\Upsilon$ are transformed in $\infty,A^*,B^*,C^*,\Upsilon$ so that the cross ratios of any four points tuple are preserved.
- The violet big circle of radius $R$ is transformed in the line through $\Upsilon=\Upsilon^*$ which is perpendicular on the diameter $\Omega\Psi$. (So the arc $\overset\frown{\Omega\Upsilon}$ through $\Omega,S,T,U,\Upsilon$ is mapped in a half-line $\infty S^* T^*U^*\Upsilon $.
- Which is the angle $\bbox[orchid]{2y}$ between the two lines $\Omega ABC\Upsilon C^*B^*A^*$ and $\Upsilon U^*T^*S^*$? We have
$$
\sin \bbox[orchid]{2y}=\sin \widehat{\Upsilon\Psi\Omega}=\frac{\Omega\Upsilon}{\Omega\Psi}
=\frac{4\sqrt{\sqrt 5-2}}2=2\sqrt{\sqrt 5-2}\ .
$$Here, we can easily compute a trigonometric function of $2y$, however we will use in the sequel expresions in $y$, the angle between the orange line of the centers and either of $\Upsilon A^*$ or $\Upsilon S^*$. Let us compute $\bbox[lightblue]{s:=\sin y}$ by solving $2s\sqrt{1-s^2}=2\sin y\cos y=\sin 2y=2\sqrt{\sqrt 5-2}$. After dividing by $2$ and squaring, $s^2(1-s^2)=\sqrt 5-2$. So $0=s^4-s^2+\sqrt 5-2$, i.e.
$$
s^2=\frac 12(1\pm \sqrt{1-4\sqrt 5 + 8})=\frac 12(1\pm(\sqrt 5-2))\ ,
$$
and we pick the value with minus, to get an angle less $45^\circ$, so
$s^2=G=\frac 12(3-\sqrt 5)=\frac 14(6-2\sqrt 5)=\left(\frac 12(\sqrt 5-1)^2\right)$. This gives:
$$
\bbox[lightblue]{
\ s=\sin y=\frac 12(\sqrt 5-1)\
}\ .
$$
There must be a geometric argument, which i am missing now, it would simplify considerably the whole exposition.
- The given green, blue, red circles have a complicated placement.
However, their inversive transforms are in a similarity, being tangent to
the two lines through $\Psi$. Let us denote by $G_i,b_i,r_i$, respectively $G_p,b_p,r_p$ the radius lengths of the transformed circles, respectively the projection of these lengths from the common orange line of their centers on either of the lines $\Upsilon C^*B^*A^*$ or $\Upsilon U^*T^*S^*$. There are
also a similarity relations for the many pairs of circles, when seen from $\Omega$.
So we have
$$
\begin{aligned}
G &=\frac{\Omega A}{\Omega A^*}G_i\ ,&
b &=\frac{\Omega B}{\Omega B^*}b_i\ ,&
r &=\frac{\Omega C}{\Omega C^*}r_i\ ,
\end{aligned}
$$
and
$\displaystyle\frac{G_p}{G_i} = \frac{b_p}{b_i} = \frac{r_p}{r_i} = \cos y$,
and
$\displaystyle\frac{G_i}{\Upsilon A^*} = \frac{b_i}{\Upsilon B^*} = \frac{r_i}{\Upsilon C^*} = \tan y$.
Considering $\Upsilon A^*$, $G_p$ as known, we determine the "next level" from the system
$$
\left\{
\begin{aligned}
\Upsilon B^*+b_p&=\Upsilon A^*-G_p
\ ,\\
\frac
{b_p}{G_p} &=\frac{b_i}{G_i}=\frac{\Upsilon B^*}{\Upsilon A^*}\ ,
\end{aligned}
\right.
$$
so we insert $b_p=\frac{\Upsilon B^*}{\Upsilon A^*}G_p$
from the second equation into the first one, getting:
$$
\Upsilon B^*
=\Upsilon A^*\cdot\frac{\Upsilon A^*-G_p}{\Upsilon A^*+G_p}
=\Upsilon A^*\cdot\underbrace{\bbox[lightblue]{\frac{1-\sin y}{1+\sin y}}}_{\bbox[lightblue]{\text{Notation: }k }}
=\Upsilon A^*\cdot \underbrace{\bbox[lightblue]{\ (\sqrt 5-2)\ }}_{\bbox[lightblue]{\ =k\ }}\ ,
$$
and for $b_p$, $b_i$ we have the same factor of proportionality:
$$
b_p = G_p\cdot k\ ,\ b_i=G_i\cdot k\ .
$$
We can now conclude:
$$
\begin{aligned}
\frac rG
&=
\frac
{\displaystyle \frac{\Omega C}{\Omega C^*}\cdot r_i}
{\displaystyle \frac{\Omega A}{\Omega A^*}\cdot G_i}
=
\frac
{\Omega C\cdot\Omega C^*}
{\Omega C^*\cdot\Omega C^*}
\cdot\underbrace{
\frac{\Omega A^*}{\Omega A}}_{=4}
\cdot\underbrace{
\frac{r_i}{G_i}}_{=k^2}
=
\frac{\Omega\Upsilon^2}{{\Omega C^*}^2}\cdot 4k^2
=
\left(\frac{\Omega\Upsilon}{\Omega C^*}\right)^2\cdot 4k^2
\\
&=
\left(\frac{\Upsilon A^*}{\Upsilon A^* +\Upsilon C^*}\right)^2\cdot 4k^2
=
\left(\frac{1}{
\displaystyle 1+ \frac{\Upsilon C^*}{\Upsilon A^*}}\right)^2\cdot 4k^2
=
\left(\frac 1{1+k^2}\right)^2\cdot 4k^2
\\
&=
\left(\frac {2k}{1+k^2}\right)^2
%=
%\left(\frac {2(\sqrt 5-2)}{1+(9-2\sqrt 5)}\right)^2
=
\left(\frac {2(\sqrt 5-2)}{10-2\sqrt 5}\right)^2
=
\left(\frac {2(\sqrt 5-2)}{2\sqrt 5(\sqrt 5-2)}\right)^2
\\
&=
\left(\frac 1{\sqrt 5}\right)^2
=\bbox[yellow]{\ \frac 15\ }\ .
\end{aligned}
$$
$\square$
Optical check:
I insisted against my will to do from the start an exact picture.
Now it partially pays back the cost of the time.
Let us see if the result matches the exact picture. We magnify the last picture at the right place...
And yes, the parallel to $\Omega\Upsilon$ drawn through the center of the red circle was used to move the red radius $r$ onto a fraction of the green radius $G$. Then taking successive reflections there is a match after the fifth step.
Mission accomplished.
LATER EDIT: (Just the story to the solution, and a bonus generalization.)
As mentioned in the comments, this solution grew out from the first picture with the big circle, and the position of the three green disks. Having this we can remove two of the disks and keep only one of them, the one that will get an appended tangent blue disk, which will be also accompanied on the other side by the tangent red disk. My initial intention was then to construct these two further disks geometrically, and check the proportionality red vs. green. (Somehow i doubted there will be a rational number in there as result, instead of a number in the field $\Bbb Q(\sqrt 5)$, i saw no (Galois) reason for such a coincidence, and i wanted to see if there is a coincidence, and if yes, then why. This was my question inside the question.) Now the geometric construction is easy and natural, use inversion to go
- from the $D$-shape of the region where all five, well only three needed disks (green, blue, red) live in,
- to an angular shape.
The transformed circles will be now similar to each other, and we need to get once for all times the factor of similarity, which is determined by the angular shape. Furthermore, each transformed disk is similar to the initial, original disk. Using this geometric argument, i could quickly construct in geogebra one by one
- the transformed green disk, tangent to the legs of the angle in $\Upsilon$,
- the "orange" angle bisector from $\Upsilon$ which will carry now the centers of the next disks,
- the centers of the next transformed disks, blue and red, and their diameters on the orange ray, one by one in geometric progression,
- and finally the initial disks.
And indeed, the red radius was i proportion $1:5$ to the green radius. I decided to make the computation, which is dictated by the construction. Going cartesian is easier, when the red disk is the final target, but the idea cannot be generalized. What if we are adding some two more circles in brown, and then in pink to the green-blue-red chain? So i did the calculus, it is always simple to do it on paper, but then name notations and display formulas...
For short, the main merit of this approach is the possibility to immediately generalize. So assume we use general notations in the following spirit:
$$
\begin{array}{|r|l|r|l|c|r|l||}
\hline
\scriptstyle\text{Disk} & &
\scriptstyle\text{Its radius} & &
\scriptstyle\text{Disk}^* &
\scriptstyle\text{Its radius} & &
\scriptstyle\text{Point on } \Omega\Upsilon &
\\\hline
%
\text{green} & = \Gamma(0) &
G & = r(0) &
\Gamma(0)^* &
G_i & = r_i(0) &
A & = A(0)
\\\hline
%
\text{blue} & = \Gamma(1) &
b & = r(1) &
\Gamma(0)^* &
b_i & = r_i(1) &
B & = A(1)
\\\hline
%
\text{red} & = \Gamma(2) &
r & = r(2) &
\Gamma(2)^* &
r_i & = r_i(2) &
C & = A(2)
\\\hline
%
\text{brown} & = \Gamma(3) &
br & = r(3) &
\Gamma(3)^* &
br_i & = r_i(3) &
D & = A(3)
\\\hline
%
\text{pink} & = \Gamma(4) &
p & = r(4) &
\Gamma(4)^* &
p_i & = r_i(4) &
E & = A(4)
\\\hline
%
\vdots & \ \vdots &
\vdots & \ \vdots &
\vdots &
\vdots & \ \vdots &
\vdots & \ \vdots
\\\hline
%
\text{?} & = \Gamma(N) &
? & = r(N) &
\Gamma(4)^* &
?_i & = r_i(N) &
? & = A(N)
\\\hline
%
\vdots & \ \vdots &
\vdots & \ \vdots &
\vdots &
\vdots & \ \vdots &
\vdots & \ \vdots
\end{array}
$$
Then the natural question is what is the proportion :
$$
\bbox[lightyellow]{\ \frac {r(N)}{r(0)} \ }
$$
for a general $N\ge 0$?
Well, the same computation delivers, using the more general notations:
$$
\begin{aligned}
\bbox[lightyellow]{\ \frac {r(N)}{r(0)} \ }
&=
\frac
{\displaystyle \frac{\Omega A(N)}{\Omega A(N)^*}\cdot r_i(N)}
{\displaystyle \frac{\Omega A(0)}{\Omega A(0)^*}\cdot r_i(0)}
=
\frac
{\Omega A(N)\cdot\Omega A(N)^*}
{\Omega A(N)^*\cdot\Omega A(N)^*}
\cdot\underbrace{
\frac{\Omega A(0)^*}{\Omega A(0)}}_{=4}
\cdot\underbrace{
\frac{r_i(N)}{r_i(0)}}_{=k^N}
\\
&=
\frac{\Omega\Upsilon^2}{{\Omega A(N)^*}^2}\cdot 4k^N
=
\left(\frac{\Omega\Upsilon}{\Omega A(N)^*}\right)^2\cdot 4k^N
\\
&=
\left(\frac{\Upsilon A(0)^*}{\Upsilon A(0)^* +\Upsilon A(N)^*}\right)^2\cdot 4k^N
=
\left(\frac{1}{
\displaystyle 1+ \frac{\Upsilon A(N)^*}{\Upsilon A(0)^*}}\right)^2\cdot 4k^N
\\
&=
\left(\frac 1{1+k^N}\right)^2\cdot 4k^N
=\bbox[lightyellow]{\ \frac{4k^N}{(1+k^N)^2}\ }\ .
\end{aligned}
$$
Recall the value:
$$
\bbox[lightblue]{\ k = \sqrt 5-2\ }\ .
$$
Now for even values of $N$, the above expression is Galois-invariant w.r.t. the conjugation (denoted by a bar) $a+b\sqrt 5\to
\overline{(a+b\sqrt 5)}:=a-b\sqrt 5$, $a,b\in\Bbb Q$, since
$k\cdot \bar k= (-k)\overline{(-k)}=(2-\sqrt 5)(2+\sqrt 5)=4-5=-1$, i.e. $\bar k=-1/k$, so $\bar k^2=1/k^2$, and thus for an even $N$,
since conjugation is a field automorphism (i.e. compatible with all field operations),
$$
\overline{\
\left(\frac{4k^N}{(1+k^N)^2}\right)
\ }
=
\left(\frac{4(1/k)^N}{(1+(1/k)^N)^2}\right)
=
\frac{4k^N}{(1+k^N)^2}\ .
$$
So the fraction is rational! (This was may question inside the question, we win rationally, since the involved polynomials are reciprocal in some sense.)
In fact, taking the inverse, $r(0):r(N)$ for an even $N=2n$, since the norm of $k^2$ is one, we get an integer!
$$
\frac{(1+k^N)^2}{4k^N}
=\frac 14(k^{n} + k^{-n})^2
=\left( \frac 12(\sqrt 5-2)^{n} + (\sqrt 5+2)^n\right)^2
\ .
$$
(And this integer is five times a square for $n=N/2$ odd, and a true
square for $n=N/2$ even. Use binomial expansion above to obtain this information.)
OK, then let us plot the first few values for the proportion
which is $>1$, for $N$ even we see no denominators:
$$
\begin{array}{|r|c|c|}
\hline
N & r(0):r(N) & \text{Factorization of } r(0):r(N)\\\hline
0 & 1 & 1\\\hline
1 & \frac{1}{2} \sqrt{5} + \frac{1}{2} & \\\hline
2 & 5 & 5\\\hline
3 & \frac{17}{2} \sqrt{5} + \frac{1}{2} & \\\hline
4 & 81 & 3^{4}\\\hline
5 & \frac{305}{2} \sqrt{5} + \frac{1}{2} & \\\hline
6 & 1445 & 5 \cdot 17^{2}\\\hline
7 & \frac{5473}{2} \sqrt{5} + \frac{1}{2} & \\\hline
8 & 25921 & 7^{2} \cdot 23^{2}\\\hline
9 & \frac{98209}{2} \sqrt{5} + \frac{1}{2} & \\\hline
10 & 465125 & 5^{3} \cdot 61^{2}\\\hline
11 & \frac{1762289}{2} \sqrt{5} + \frac{1}{2} & \\\hline
12 & 8346321 & 3^{6} \cdot 107^{2}\\\hline
13 & \frac{31622993}{2} \sqrt{5} + \frac{1}{2} & \\\hline
14 & 149768645 & 5 \cdot 13^{2} \cdot 421^{2}\\\hline
15 & \frac{567451585}{2} \sqrt{5} + \frac{1}{2} & \\\hline
16 & 2687489281 & 47^{2} \cdot 1103^{2}\\\hline
17 & \frac{10182505537}{2} \sqrt{5} + \frac{1}{2} & \\\hline
18 & 48225038405 & 5 \cdot 17^{2} \cdot 53^{2} \cdot 109^{2}\\\hline
19 & \frac{182717648081}{2} \sqrt{5} + \frac{1}{2} & \\\hline
20 & 865363202001 & 3^{4} \cdot 41^{2} \cdot 2521^{2}\\\hline
\end{array}
$$
(... and we may also want to say something about the case of an odd $N$?!)
The above table was easily produced:
K.<a> = QuadraticField(5, latex_name=r'\sqrt{5}')
k = a - 2
for N in [0..20]:
f = (1 + k^N)^2 / (4*k^N)
print(f"{N} & {latex(f)} & "
f"{latex(QQ(f).factor()) if f in QQ else ''}"
r"\\\hline")
So, to have a final short message, if we further add to the green-blue-red-chain a further brown-pink appendix, then the pink radius is $1/81$
of the green one.
