Unique Tangency of circles from Sangaku A Geometric Sangaku shape here on this wooden board up right from Japan's Buddhist Temples in 1859 during Edo period drew my attention recently, however I've found the exact values of the radius of all the circles by using inversion transformation and solving some algebraic equations by machine, It is still remained wonders on how they draw or craft this shape on the board such precisely those days.
 
the ratio of radiuses of two big circle is:
$\frac{\overline{OL}}{\overline{OC}}=\frac{2\sqrt{7}\cos(\frac{\arccos(\frac{\sqrt{7}}{14})}{3})-1}{3} \approx 1.246979603717467$
if we take the radius of the smaller circle $\overline{BJ}=1$ then the radiuses of two big circles are: $\overline{OL}\approx 3.603875471609681,\overline{OC}\approx 2.890083735825261$
I used inversion with the center of $O$, the blue circle as inversion circle with power $\overline{OL} ^2$ and used equations to find relation of radius of the little orange circle (which is the inverse of the maroon circle with center of $G$) with other circle's radii.
In quest for an absolute geometric way of finding radii relations and drawing the whole shape with compass and ruler ,I've encountered another problem which seems irrelevant to this problem, but it would shows another complex aspect of tangent circles on plain.

Question: Can anyone find an absolute geometric solution to find
radii relations and drawing the whole shape with compass and ruler?
hint: from my solution it came out that yAxis is tangent to the orange
circle on the center of green circle $C$.

 A: 
Writing $a$, $b$, $c$ for the radii of the large, medium, and small circles, respectively, we identify two key triangles in the figure.

*

*A right triangle with legs $b$ and $a-c$, and hypotenuse $b+c$. By Pythagoras,
$$b^2+(a-c)^2=(b+c)^2 \quad\to\quad c = \frac{a^2}{2(a+b)} \tag1$$


*A triangle with sides $b$, $a+c$, $b-c$, and altitude $c$ relative to base $b$; expressing the square of the area in two ways (the second being Heron's formula), we have
$$\begin{align}\left(\tfrac12bc\right)^2=\tfrac1{16}(b+(a+c)+(b-c))&\cdot(-b+(a+c)+(b-c))\\ &\cdot(\phantom{-}b-(a+c)+(b-c)) \\
&\cdot(\phantom{-}b+(a+c)-(b-c))
\end{align} \tag2$$
Substituting from $(1)$ into $(2)$ eliminates $c$; leaving (for $a>0$) this relation:
$$a^3 + a^2 b - 2 a b^2 - b^3= 0 \tag3$$ which is consistent with OP's calculated values $a=3.60387\ldots$ and $b = 2.89008\ldots$.
(Alternatively, eliminating $b$ gives $a^3 - 2 a^2 c - 8 a c^2 + 8 c^3= 0$, and eliminating $a$ gives $b^3 - 8 b^2 c + 12 b c^2 + 8 c^3= 0$. The latter confirms @SangchulLee's comment under the question.)
Because $(3)$ is an irreducible cubic, we conclude that a straightedge-and-compass construction of this figure is impossible.
That said, I see no reason to believe that the creator of a sangaku was in any way restricted in the tools or techniques used to draw the figure. I don't doubt that numerical approximations were allowed and/or expected (no real-world diagram has infinite precision, after all). That said, there are ways to extract, say, cube roots with the help of a marked ruler, so the figure in question may well be geometrically "constructible" in an appropriate sense of the term.
I'd be interested to learn if there were specific geometric recipes for the kinds of computations implicit in sangaku figures. I recommend OP post a separate question to address this issue. The History of Science and Mathematics StackExchange may be able to provide historical context.
