= Solution
Put $w=1/z$, so $z=1/w$ and $w\ne0$. Substitute into the equation in part (a) and multiply by $|w|^2$:
$$
\frac{A}{|w|^2}+\frac{\overline B}{w}+\frac B{\overline w}+C=0
\quad\Longleftrightarrow\quad
\boxed{C|w|^2+Bw+\overline B\overline w+A=0.}
$$
This is the <reciprocal transformation of a generalized circle>. In the original coefficient convention the new triple is $(C,\overline B,A)$: the linear coefficient is conjugated, which reflects the distinction between $1/z$ and <inversion in a circle> $1/\overline z$.
\b[A <circle> through zero.] Here $A\ne0$, $C=0$, and nondegeneracy forces $|B|^2>0$. Its image satisfies
$$
2\operatorname{Re}(Bw)+A=0.
$$
This is a <straight line>, and $w=0$ does not satisfy it because $A\ne0$. Conversely every point of this <straight line> has a nonzero reciprocal on the original <circle>. Thus the original <circle> with zero omitted maps onto the full finite <straight line>.
\b[A <circle> missing zero.] Now $A\ne0$, $C\ne0$ and $\Delta=|B|^2-AC>0$. Applying part (a) to the transformed coefficients gives centre $-\overline B/C$ and <radius> $\sqrt\Delta/|C|$, so the image is again a genuine <circle>. It misses zero, because the transformed equation has nonzero constant $A$. The inverse $z=1/w$ proves that the whole of this <circle> is attained.
\b[A <straight line> through zero.] Here $A=C=0$ and $B\ne0$. The transformed equation is $2\operatorname{Re}(Bw)=0$, another <straight line> through zero. Precisely, the finite map takes the punctured original <straight line> onto the punctured image <straight line>, since neither the domain nor the image of $1/z$ contains zero.
In the <Riemann sphere> convention $0\leftrightarrow\infty$, all three become exact statements about whole <generalized circles>:
$$
\boxed{\begin{aligned}
\text{circle through }0&\longmapsto\text{line missing }0,\\
\text{circle missing }0&\longmapsto\text{circle missing }0,\\
\text{line through }0&\longmapsto\text{line through }0.
\end{aligned}}
$$
Finite <straight lines> acquire their point at infinity in this convention. Stating it avoids accidentally adding $w=0$ merely because it satisfies a multiplied equation.
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