Sangaku. How to draw those three circles with only a ruler and a compass? I found in a book of Sangakus the following problem.

Let $R_b$, $R_g$ and $R_r$ the radiuses of the blue, green and red circles $C_b$, $C_g$ and $C_r$.
Prove that
$$\frac{1}{\sqrt{R_r}}=\frac{1}{\sqrt{R_b}}+\frac{1}{\sqrt{R_g}}\,.\quad (1)$$
And this I can do. But then
I would like to draw the figure myself with only a ruler and a compass.
I know it is possible as I can construct inverses, sums and square roots with only a ruler and a compass, but when I tried to draw the figure with a "simple" or "natural" construction, I failed.
Does someone have an idea of how to draw it "naturally"?

EDIT
Answer to a comment of Amzoti:
To prove Relation (1) I first prove the relation:
$$AB^2=4R_gR_b\quad (2)$$

Relation (2) is a consequence of Pythagoras' theorem in the triangle $O_bO_gH$:
$$AB^2+(R_g-R_b)^2=(R_b+R_g)^2\,.$$
(It was the previous Sengaku in the book.)
We thus get the relations
$$
\begin{align}
AB^2 & =4R_gR_b\quad (2) \\
AC^2 & =4R_gR_r\quad (3) \\
BC^2 & =4R_bR_r\quad (4) \\
\end{align}
$$
where $A$, $B$, $C$ are the orthogonal projections of the centers of the circles $C_b$, $C_g$ and $C_r$ on the line $d$. The relation $AB^2=AC^2+BC^2+2AC.BC$ then yields, using Relations (2) to (4),
$$4R_gR_b=4R_gR_r+4R_bR_r+8\sqrt{R_bR_g}R_r$$
Divided by $4R_rR_gR_b$ this equation is
$$\frac{1}{R_r}=\Big(\frac{\sqrt{R_g}+\sqrt{R_b}}{\sqrt{R_bR_g}}\Big)^2$$
which is in fact Relation (1) squared.
 A: 
I found another straightforward method which can be perfectly show the meaning!
A: We'll solve the following problem:
Given, a straight line $r$, a point $A$ such that $A \in r$ and the radii $R_b$ and $R_g$, construct three circles $\lambda_b$,$\lambda_g$ and $\lambda_r$, such that they are tangent to each other and to $r$, and $R_r \leq min(R_b, R_g)$.($R_b$, $R_g$, and $R_r$ are the radii of $\lambda_b$,$\lambda_g$ and $\lambda_r$ respectively).
See the figure below:



*

*Draw a straight line $t$ such that $A \in t$ and $ t \perp r$.

*Mark $O_b$ such that $O_b \in t$ and $O_bA = R_b$.

*Draw $\lambda_b$.

*Draw $s$ such that $s \parallel r$ and $d(s,r)=R_g$.

*Draw an arc $\mu$ centered at $O_b$ and radius  $R_b + R_g$ such that $\mu$ intesects $s$. The intersection point is $O_g$.

*Draw $\lambda_g$.

*Find out $R_r$. (Auxiliary construction).

*Find out $O_r$. ($O_r$ is the intersection point of two arcs: one of them has $O_b$ as center and $R_b + R_r$ as radius and the other one has $O_g$ as center and $R_g + R_r$ as radius).

*Draw $\lambda_r$.


Auxiliary construction 
Let $d=d(A, B)$ , where $\{B\}= r \cap \lambda_g$.


*

*From a point $Q$ draw two arbitrary rays $w$ and $h$.

*Mark a point $N$ such that $d(N,Q) = R_b + d + R_g$ and $N \in h$.

*Mark a point $M$ such that $d(M,Q) = R_b$ and $M \in w$.

*Draw a straight line $u$ such that $u$ passes through $M$ and $N$.

*Mark a point $P$ such that $d(P,Q) = R_b + d + R_g + R_g $ and $P \in h$.

*Draw a straight line $v$ such that $v \parallel u$ and $v$ passes through $P$.

*$R_r = d(M,T)$ where $\{T \}= v \cap u$.

A: Perhaps see section 9.3 of Alter Mundus by Alain Matthes
Section 9 also contains a proof of your stated problem and it would be nice to compare to your proof if you would be so kind to add it to your question.
It looks like the web site is still under construction, but also has some items of interest.
I have not tried the construction in 9.3 above, but and am curious if it works.
Update: by the way, you might also want to check out some software and there are MSE recommendations on hyperbolic-geometry-software-programs listed here.
Regards
A: Perhaps look in the Geometry Junkyard
http://www.ics.uci.edu/~eppstein/junkyard/tangencies/apollonian.html
A: 
I think my method is the one you may want. it is nature.
it is not difficult to proof. just care $KE=\sqrt{r1*r2}$ and KM=MB
