 Solution

Put $w=1/z$, so $z=1/w$ and $w\ne 0$. Substitute into the equation in part (a) and multiply by $|w{|}^2$: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 $1/\overline{z}$.

A circle through zero. Here $A\ne 0$, $C=0$, and nondegeneracy forces $|B{|}^2>0$. Its image satisfiesThis is a straight line, and $w=0$ does not satisfy it because $A\ne 0$. 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.

A circle missing zero. Now $A\ne 0$, $C\ne 0$ 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.

A straight line through zero. Here $A=C=0$ and $B\ne 0$. The transformed equation is $2\mathrm{Re}(Bw)=0$, 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: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.