= Solution
For the reciprocal <Möbius transformation>, substitute $z=1/w$, $\bar z=1/\bar w$ and multiply by $w\bar w$, which is nonzero on the stipulated domain. This yields
$$
a+bw+\bar b\bar w+cw\bar w=0.
$$
Consequently
$$
\boxed{A=c,\qquad B=\bar b,\qquad C=a},
\qquad
\boxed{A=0\iff c=0}.
$$
For a nondegenerate original <circle>, $c=0$ says that it passes through the origin, and its image is a <straight line>. The restriction $w\ne0$ must be retained: multiplying through does not authorize adding that excluded point. Degenerate choices can give a constant or an identically zero equation instead of a <straight line>.
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