In a composition of two inversions, characterize straight lines that remain straight lines Let S1 and S2 be the inversions in the circles $x^2+y^2 = 16$ and $(x−8)^2+y^2 =
1$ respectively. Consider the composition W for these inversions W = T2 ◦T1. Which lines
under the transformation W will become lines?
From my knowledge, I know that lines are inverted to lines when they pass through the center of the circles of inversion. So for the line to remain lines under inversion does the non-inverted line have to pass through the centers of both the circles above? Is that the only way it remains a line?

Circles and lines mapped to lines by circle inversion are exactly those which pass through the center of inversion; the center is the only point mapped to infinity. However, inversion is a bijection so there is exactly one point that $S_1$ maps to the center of $S_2$. It is the point on the line between the centers of $S_1$ and $S_2$ a distance $d$ from the center of $S_1$ such that
$$
  8d = 16 \implies d = 2
$$
since the center of $S_2$ is 8 units away from the center of $S_1$ and 16 is the squared radius of $S_1$. This point has coordinates $(2, 0)$. The set of all lines which map to lines under $S_2\circ S_1$ is exactly all lines which pass through this point. Note that the radius of $S_2$ is irrelevant.
