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First normalize the two circles by translation and rotation so their centres are $0,d$ and radii $a,b$. If $d=0$ nothing further is needed. Otherwise choose real $s,t$ satisfyingThe discriminant of this quadratic issince two disjoint circle boundaries are either externally separated or strictly nested. Thus $s\ne t$, and $M(z)=(z-s)/(z-t)$ is a Möbius transformation. Direct expansion shows that on the first circle $|M(z){|}^2=s/t$, and on the second $|M(z){|}^2=(s-d)/(t-d)$. Both constants are positive; the pole is on neither circle. Their images are distinct concentric circles. This proves the concentric normalization of disjoint circles.

For a construction with $n\ge 3$, use inner radius $\rho =1$, outer radiusand small circles of radius $r$ centred at $d{e}^{2\pi ij/n}$. Each touches the two boundaries, and adjacent centre distances are $2d\sin (\pi /n)=2r$. All other distances are at least $2r$, so there are no unwanted intersections. These are Steiner chains. For the literal $n=2$ requirement, take $\rho =1$, $R=3$, and two radius-one circles centred at $2e^{\pm i\pi /6}$. Their centre distance is $2$, so this gives two distinct mutually tangent circles touching both boundaries. Existence holds for every $n\ge 2$, but the two-circle case has only one tangency between neighbours, repeated by the cyclic indexing.

For an annular circle touching radii $\rho ,R$, its radius and centre distance are $r=(R-\rho )/2$ and $d=(R+\rho )/2$. Two neighbouring such circles have angular separation $\delta =2\arcsin (r/d)$; their tangency point is the midpoint of their centres. Its distance from the common centre isThus the tangency locus in concentric coordinates is a circle of radius $\sqrt{R\rho }$. For $n\ge 3$, disjointness forces the constellation to use these annular circles. Indeed, the other family of circles touching both boundaries has radius $d$ and centre distance $r$, enclosing the inner boundary. Two members of that larger family intersect. One such circle can be disjoint from an annular circle only at the two opposite centre directions where their boundaries are tangent; those two annular circles cannot be tangent to each other. Hence this family cannot occur in a constellation of three or more circles.

Under the inverse Möbius transformation, the tangency locus is a generalized circle, which can be an ordinary circle or a straight line. The printed claim that it is always an ordinary circle is false without this qualification. For an explicit counterexample, use the above $n=3$ construction, put $L=\sqrt{R}$, and apply $w=1/(z-L)$. Its pole lies on the tangency locus but on none of the original boundary or chain circles. Consequently all the transformed individual circles remain ordinary circles. The three distinct tangency points transform to three points on the line $\mathrm{Re}w=-1/(2L)$, so they cannot lie on an ordinary circle.

The usual closure assertion is Steiner's porism, with two essential conventions: stay in the annular family and always choose the next tangent circle in the same angular direction. For a disjoint closed chain with $n\ge 3$, distinctness prevents reversing direction: a reversed step would return immediately to the preceding circle. Thus every step has the same sign and closure gives $n\delta =2\pi m$ for an integer $m\ge 1$. If $m>1$, the $n$ distinct centres have some angular gap $2\pi /n<\delta$, making two circles intersect. Hence $m=1$ and $\delta =2\pi /n$. Starting at any new angle ${\theta }_0$ produces centres $d{e}^{i({\theta }_0+j\delta )}$, a rotation of the original chain, soMapping back proves the same porism on the corresponding branch of circles for the original pair.

The literal final request, with arbitrary tangent choices and $n=2$ included, is stronger and false. In the two-circle example above, the forward angular step is $\pi /3$. Starting at angle $0$, a next circle at $\pi /3$ and then one at $2\pi /3$ satisfy every inductive tangency requirement, but $Y_2\ne Y_0$. Even for $n\ge 3$, allowing backtracking lets the construction reverse instead of completing the chain. Thus the genuine geometric conclusion is the qualified porism, not unrestricted closure under the printed induction.