= Solution
\b[The three contacts in the strip.] Work with the external-contact arrangement drawn in the original diagram: both <disks> lie between the two <parallel lines> and touch opposite boundaries. A <rigid motion> puts those boundaries at $x=0$ and $x=d$, with $d>0$, and makes $P=(0,0)$. Write $R=(d,h)$, and let the positive <radii> be $r_1,r_2$. Because a <tangent to a circle> is perpendicular to the corresponding <radius>, the centres are
$$
O_1=(r_1,0),\qquad O_2=(d-r_2,h).
$$
At the <external tangency>, the two radial segments form the segment $O_1O_2$, with $|O_1Q|=r_1$ and $|QO_2|=r_2$. Hence the internal division formula gives
$$
Q=\frac{r_2O_1+r_1O_2}{r_1+r_2}=\left(\frac{r_1d}{r_1+r_2},\frac{r_1h}{r_1+r_2}\right)=\frac{r_1}{r_1+r_2}R.
$$
Thus
$$
\boxed{Q=P+\frac{r_1}{r_1+r_2}(R-P),\qquad P,Q,R\text{ are collinear}.}
$$
The coefficient is strictly between zero and one, so $Q$ lies on the actual <line segment> $PR$. This proves <collinearity for tangent circles between parallel lines>. We will also use the fact that $PQ$ and $QR$ are <chords> of the respective <circles>; the entire <line segment> $PR$ therefore lies in the union of their closed <disks>.
\b[The four contacts.] Label the successive contact points $T_{12},T_{23},T_{34},T_{41}$, with $T_{jk}$ belonging to $C_j$ and $C_k$. The original diagram has four positive-<radius> <circles> with disjoint interiors, successive <external tangencies>, and distinct contact points. Set $T=T_{41}$ and apply the reciprocal <Möbius transformation>
$$
w=\frac1{z-T}.
$$
By part (b), the two <circles> $C_1,C_4$ through $T$ become <straight lines> $\ell_1,\ell_4$. These <straight lines> cannot meet at a finite point, since their inverse images would give a second intersection of $C_1,C_4$. They are therefore distinct <parallel lines>. The other two <circles> avoid $T$, so their images $D_2,D_3$ are ordinary <circles> missing zero.
The three remaining contacts retain their tangencies. Indeed, away from $T$ the map and its inverse are <holomorphic functions> with nonzero <derivatives>, hence are local <conformal maps>. In particular they preserve a shared <tangent line>. Equivalently, bijectivity preserves the unique intersection of two <generalized circles>, which characterizes their tangency here.
The finite closed <disks> bounded by $D_2,D_3$ lie between $\ell_1,\ell_4$, have <external tangency> to each other, and touch opposite boundaries at the images of $T_{12}$ and $T_{34}$. To track the sides as well as the boundaries, a closed <disk> avoiding $T$ maps to the bounded closed <disk> of its image <circle>, while each <disk> with $T$ on its boundary maps to a closed <half-plane>. The two latter <half-planes> have disjoint interiors, so their complement is the strip containing zero. Disjointness of the original interiors puts $D_2,D_3$ inside that strip. Neither image closed <disk> contains zero, since the reciprocal <Möbius transformation> sends only infinity to zero.
The first part now makes the three finite transformed contact points <collinear>, on a <straight line> $\ell$. It is essential that $\ell$ misses zero. If it passed through zero, its segment between $\ell_1$ and $\ell_4$ would contain zero, since zero lies strictly within the strip. But the first part shows that this whole <line segment> is covered by the two image closed <disks>, neither of which contains zero. This contradiction proves $0\notin\ell$.
The inverse image of a <straight line> missing zero is a genuine <circle> through $T$, by the same coefficient transformation in part (b), or by applying the involution $u\mapsto1/u$ in reverse and translating back. This <circle> contains $T_{12},T_{23},T_{34}$, while $T$ is the inverse image of the point at infinity on $\ell$. Therefore
$$
\boxed{T_{12},\ T_{23},\ T_{34},\ T_{41}\text{ lie on one circle}.}
$$
This is the <four contact points of a circle chain> result; the four contacts form a <cyclic quadrilateral>.
\Image[/past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/metric-2015-tangency-inversion.png]
{title=The <reciprocal transformation of a generalized circle> turns the chain into two <tangent circles> between <parallel lines>}
{description=An original exact construction illustrates the arrangement. The orange <circle> through all four contacts becomes the orange <straight line> through the three finite transformed contacts. The hollow point at zero lies between the <parallel lines> and outside the two image <disks>.}
The pictured arrangement matters. Arbitrary mixtures of <internal tangency> and <external tangency> can instead give four distinct <collinear> contacts: take four <circles> whose diameter endpoints on the <real axis> are respectively $[0,3]$, $[0,1]$, $[1,2]$, and $[2,3]$. Their successive contacts are $0,1,2,3$, so no ordinary <circle> contains all four. The proof above establishes the requested genuine-<circle> conclusion for the external-contact arrangement actually drawn, rather than silently treating every <generalized circle> as an ordinary <circle>.
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