= Collinearity for tangent circles between parallel lines
Let two positive-<radius> <circles> have <external tangency> at $Q$ and lie in the closed strip between two <parallel lines>, touching opposite boundaries at $P,R$. Then $P,Q,R$ are <collinear>. To prove this, choose coordinates with the boundaries $x=0,d$, put $P=(0,0)$ and $R=(d,h)$, and write the centres as $O_1=(r_1,0)$, $O_2=(d-r_2,h)$. The contact point is
$$
Q=\frac{r_2O_1+r_1O_2}{r_1+r_2}=\frac{r_1}{r_1+r_2}R.
$$
The closed <line segment> $PR$ is contained in the union of the two closed <disks>: its portions from $P$ to $Q$ and from $Q$ to $R$ are <chords>. The strip and <external tangency> hypotheses are essential to this particular formulation.
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