In the area between the two circles, draw a circle K

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 C_{1} and C_{2}.

All the circles orthogonal to both circles are centered on Δ
and intersect the center line C_{1}C_{2} 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 C_{1} and C_{2}.

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 C_{1}.

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 C_{1} has same radius as C'_{1},
hence C_{1} 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 C_{1}

Center M of K is intersect point of OT with IM'.

Circle C_{2} is also easily constructed from a touching point of C'_{2} with some circle K'.

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/(r

(r_{1} - r_{2})/(r_{1} + r_{2}) = (m - 1)/(m + 1) = sin(π/n)
That is : m = r _{1}/r_{2} = (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)| |

r'_{1}/r'_{2} = r_{1}/r_{2} × C_{2}(A)/C_{1}(A) |

The inversion pole A is one of the limit points A,B of the pencil of circles defined by C_{1} and C_{2}.
Circle with diameter AB is orthogonal to both circles C_{1} and C_{2}.
Hence ABM_{1}N_{1} is an harmonic range, and
C_{1}(A) = AM_{1}.AN_{1} = AB.AO_{1}
which gives C_{2}(A)/C_{1}(A) = AO_{2}/AO_{1}

Let I the midpoint of O_{1}O_{2} and d = O_{1}O_{2} distance of centers.
H having the same power to both circles,
HO_{1}² - r_{1}² = HO_{2}² - r_{2}²
that is (HO_{1} - HO_{2})(HO_{1} + HO_{2}) = 2d.HI = r_{1}² - r_{2}²

Hence HO_{1} = HI + IO_{1} = (r_{1}² - r_{2}² + d²)/(2d)

HA = HT and HT² = HO_{1}² - r_{1}² finally give
AO_{1} = HO_{1} - HA :

AO_{1} = [ r_{1}² - r_{2}² + d² - √( (r_{1}² - r_{2}² + d²)² - 4d²r_{1}²) ] / (2d) |

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