Steiner chain of circles

Given two circles C1, and C2 inside C1.
In the area between the two circles, draw a circle K1 touching both C1 and C2, then a circle K2 touching C1, C2 and K1, then K3 touching C1, C2 and K2 ... Kn touching C1, C2 and Kn-1.

Locus of centers of circles K

Distances of center of some circle K with radius ρ to centers of circles C1 and C2 are r1 - ρ and r2 + ρ, the sum is r1 + r2 then center of K lies on an ellipse whose foccii F1 and F2 are the centers of circles C1 and C2, that is the locus of points with sum of distances to F1 and F2 being constant.

Steiner chain

Sometimes, this chain closes and Kn also touches K1.
Prove that we can then choose the first circle anywhere centered on this ellipse.

Two non intersecting circles can always be transformed by an inversion into two concentric circles :
Let Δ the radical axis of the two circles C1 and C2.
All the circles orthogonal to both circles are centered on Δ and intersect the center line C1C2 in two fixed points A and B.
An inversion with pole A transforms the pencil of these orthogonal circles into a pencil of lines through B', transformed of B, orthogonal to both C'1 and C'2, transformed of circles C1 and C2.
These lines are then diameters of both C'1 and C'2. These two circles have then same center B'.
Note : We can freely choose the inversion power, for instance to keep radius of C1.

The set of circles K is then the inverse of the set of circles K', which build a Steiner chain in two concentric circles.
Of course a Steiner chain in two concentric circles can freely rotate ! QED.

This allows to construct Steiner chains with compass and straightedge :

First of all construct a Steiner chain in two concentric circles.
Of course construct first all circles K' on a regular n-gon, then C'1 around and C'2 inside.
Otherwise choosing any circles C'1 and C'2 would have no chance to fit.
Choose an inversion pole I and inversion power k. Here k is negative and the inversion circle -k is orthogonal to C'1 (IS' _|_ O'S'), so that the transformed C1 has same radius as C'1, hence C1 is constructed just by a center symmetry of C'1 through I.
Circles K are then easily constructed with just straightedge : the transformed of touching point T' is intersect point T of IT' with C1
Center M of K is intersect point of OT with IM'.
Circle C2 is also easily constructed from a touching point of C'2 with some circle K'.

Deduce a criterion for existence of such a Steiner chain

In two concentric circles, the criterion is easy :
the radius of Steiner circles is half the side of a regular n-gon, and this radius should also be half the difference of radii of the two given circles.
This gives : ρ/(r2 + ρ) = sin(π/n) and r1 = r2 + 2ρ, then substituting ρ and naming m = r1/r2 :
 (r1 - r2)/(r1 + r2) = (m - 1)/(m + 1) = sin(π/n) 
 That is :
 m = r1/r2 = (1 + sin(π/n)) / (1 - sin(π/n)) 

In the case of any two given circles, we have to calculate the ratio of radii of transformed circles in an inversion from pole one of the limit points of the pencil of circles defined by the two given circles.

First of all the relation between radii of two inverse circles.
Inversion pole is also a dilation center and AN/AM = R'/R.Inversion results into AM.AM' = k hence AN.AM' = k.R'/R But AN.AM' = C'(A) is the power of A to C'. Hence

 R'/R = |C'(A)/k| = |k/C(A)| 
This gives the ratio of radii of transformed circles from the ratio of given circles.
 r'1/r'2 = r1/r2 × C2(A)/C1(A) 

The inversion pole A is one of the limit points A,B of the pencil of circles defined by C1 and C2. Circle with diameter AB is orthogonal to both circles C1 and C2. Hence ABM1N1 is an harmonic range, and C1(A) = AM1.AN1 = AB.AO1 which gives C2(A)/C1(A) = AO2/AO1

Let I the midpoint of O1O2 and d = O1O2 distance of centers. H having the same power to both circles, HO1² - r1² = HO2² - r2² that is (HO1 - HO2)(HO1 + HO2) = 2d.HI = r1² - r2²
Hence HO1 = HI + IO1 = (r1² - r2² + d²)/(2d)

HA = HT and HT² = HO1² - r1² finally give AO1 = HO1 - HA :

 AO1 = [ r1² - r2² + d² - √( (r1² - r2² + d²)² - 4d²r1²) ] / (2d) 

We then get the criterion for existence of a Steiner chain if d ≠ 0 :

(If d → 0, the complicated term → 1)


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