= Solution
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 $s,t$ satisfying
$$
st=a^2,\qquad s+t=\frac{d^2+a^2-b^2}{d}.
$$
The discriminant of this quadratic is
$$
\frac{(d^2-(a+b)^2)(d^2-(a-b)^2)}{d^2}>0,
$$
since 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 $n\geq3$, use inner radius $\rho=1$, outer radius
$$
R=\frac{1+\sin(\pi/n)}{1-\sin(\pi/n)},\qquad r=\frac{R-1}{2},\qquad d=\frac{R+1}{2},
$$
and 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. \b[Existence holds for every $n\geq2$], but the two-<circle> case has only one tangency between neighbours, repeated by the cyclic indexing.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/ib/paper-2-circle-chains.png]
{title=Two mutually tangent circles and a five-circle Steiner chain between concentric boundaries}
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 is
$$
d\cos(\delta/2)=\sqrt{d^2-r^2}=\sqrt{R\rho}.
$$
Thus \b[the tangency locus in concentric coordinates is a <circle> of radius $\sqrt{R\rho}$]. For $n\geq3$, 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. \b[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 $\operatorname{Re}w=-1/(2L)$, 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 $n\geq3$, 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 $m\geq1$. 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 $de^{i(\theta_0+j\delta)}$, a rotation of the original chain, so
$$
\boxed{Y_n=Y_0\quad\text{for the consistently directed annular chain}.}
$$
Mapping 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 $n\geq3$, 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.
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