Chain of Inscribed Circles
Construct two tangent circles Σ_{1} and Σ_{2} and the line L through their centers. We are going to inscribe into the crescentshaped space between the circles a chain of pairwise tangent circles. The first one, α_{0}, has its diameter on L. The second, α_{1}, is tangent to α_{0}, and both Σ_{1} and Σ_{2}, and so on. Let h_{n} and R_{n} denote the distance to L from the center and the radius of the circle α_{n}. Then
(1)  h_{n} = 2n·R_{n}. 
What if applet does not run? 
Indeed, an inversion with the center at the common point of tangency of Σ_{1} and Σ_{2} maps the two circles onto two parallel lines perpendicular to L. The circles α_{n} map an a chain of equal circles inscribed between the two parallel lines with the diameter of the first one on L. For those circles, the analogue of (1) is obvious: h'_{n} = 2n·R'_{n}, where h'_{n} is the distance from the center of the inverse image of α_{n} to L and R'_{n} is its radius. However, the circles in every inverse pair are homothetic with the center of homothety at the center of inversion. Let d_{n} and d'_{n} be the distances from the center of inversion to the centers of circle α_{n} and its inverse image. Then comparing the similar triangles formed by their line of centers and the common tangent gives a proportion
d_{n}/R_{n} = d'_{n}/R'_{n}.
Now looking ad the second pair of similar triangles, viz., those formed by the center line and the horizontal axis we get
h_{n}/d_{n} = h'_{n}/d'_{n}.
Whence,
(The above proposition is commonly referred to as Pappus Circle Theorem or Pappus Chain Theorem. Pappus was of course unaware of the inversion. His proof is much more elaborate. He himself refers to it as "The Ancient Theorem" [Gaba].)
We can easily derive an additional result.
Assume the radius of the right semicircle is k times the radius of the left one. If the former is R_{R}, and the latter is R_{L}, we assume that
(2)  R_{R} = k·R_{L}. 
Let R denote the radius of the big semicircle, then, since
(3) 
R_{L} = R/(1 + k) and R_{R} = R·k/(1 + k). 
For the sake of simplicity, let's choose the radius of the circle of inversion to be 2R. Then the left vertical line (this is the one that corresponds to the big semicircle) becomes tangent to both the big and the right semicircles as the common point of tangency lies then on the circle of inversion and is thus fixed. The right vertical line corresponds to the left semicricles. Introduce further notations. Let O_{n} be the center of a_{n} and P_{n} the projection of O_{n} on L. Let AB, CB, and AC denote the diameters of the semicircles. A is the center of inversion,
(2R)² = AB² = AC·AC'.
Thus
AC' = (2R)²/AC = 4R²/2R_{L} = 2(1 + k)R.
And, since AO' = (AB + AC')/2,
AO' = (2 + k)R,
BO' = kR.
So, for the same reason as (1),
h_{n}/AP_{n} = 2nBO'/AO', or
R_{n}/AP_{n} = BO'/AO'
And
(4)  AP_{n} = AO'·R_{n}/BO' = (2 + k)R_{n}/k. 
Next, let's derive a generalization of Archimedes' formula:
(5)  R_{n} = Rk / (k²n² + k + 1). 
The points of tangency of the common tangents through A of a pair of corresponding circles (like T_{n} and T'_{n} in the diagram) play a double role. On one hand, they are homologuous, corresponding under the implied homothety. On the other hand, they are antihomologous, mapping on each other under the inversion. More accurately
AT'_{n} / AT_{n} = R'_{n} / R_{n} = Rk / R_{n}, and also
AT'_{n} · AT_{n} = (2R)²
implying
(AT'_{n})² = 4R^{3}k / R_{n}.
But (AT'_{n})² = (2R + kR)² + (2n·kR)²  (kR)² which gives an equation for R_{n}, with the solution
R_{n} = Rk / (n²k² + k + 1).
which is (5). The combination of (4) and (5) gives
(6)  AP_{n} = R(k + 2) / (k²n² + k + 1). 
Note that, t = 1/k, so that R_{L} = t·R_{R} then (5) becomes
(5')  R_{n} = Rt / (t² + t + n²). 
References
 M. G. Gaba, On a Generalization of the Arbelos, Am Math Monthly, Vol. 47, No. 1 (Jan., 1940), 1924

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