Is it necessary to consider inversion $\left(z\mapsto\dfrac{1}{z}\right)$ on the extended complex plane 
Is it necessary to consider inversion $\left(z\mapsto\dfrac{1}{z}\right)$ on extended complex plane rather than on $\mathbb C$ to map a circle passing through the origin onto a st. line?

 A: It depends on what your source intends "mapping onto" to mean.
One interpretation of "mapping $A$ onto $B$" is that every element of $B$ has an element of $A$ mapped to it--in particular, the mapping needn't be defined on all of $A$, nor must every element of $A$ be mapped to an element of $B$. In this case, you needn't worry about the extended plane.
Another interpretation is that the mapping is defined on all of $A$ and every element of $B$ is has an element of $A$ mapped to it (and possibly also that every element of $A$ is mapped to a point of $B$, so that $A$ is mapped onto and into $B$). That would require extending the plane.
A: The map does not send a line through origin to circles. 
First do not consider extended or not. Any line or circle through origin should be mapped to a line since it is not bounded by inversion map. Any line or circle not containing origin should be mapped to a circle since it is bounded by inversion map.
And considering the extended plane, the infinity point will always be mapped to origin. So any line should be mapped to line or circle containing origin. Any circle should be mapped to line or circle not containing origin.
Summary


*

*Line maps to line/circle through origin

*Circle maps to line/circle not containing origin

*Line/circle through origin maps to line

*Line/circle not containing origin maps to circle


Edit I think it does not need to consider the extended complex plane, since in usual a straight line is considered in complex plane, not considering the infinity. But it will depends on the definition what a straight line is.
