Plane intersection by a mapping and different cases In $\mathbb{R}^3$ say we have the 2 planes $A=\{z=1\}$ and $B=\{x=1\}$. A line through 0 meeting $A$ at $(x,y,1)$ meets $B$ at $(1,y/x,1/x).$ Consider the map $\phi: A \rightarrow B$ defined by $(x,y) \mapsto (y' = y/x, z' = 1/x)$.
I'm trying to figure out the image under $\phi$ of
1) the line $ax = y + b$; the pencil of parallel lines $ax = y + b$ with fixed $a$ and variable $b$;
2) circles $(x-1)^2 + y^2 = c$ for variable $c,$ distinguishing the 3 cases $c>1, c = 1,$ and $c< 1$.
and to imagine the above as a perspective drawing by an artist sitting at $(0,0,0)$ and drawing  figures from the plane $A$ on the plane $B$.
What happens to the points of the 2 planes where $\phi$ and $\phi^{-1}$ are undefined? Thanks!

To answer your last question first, notice that a line through $0$ meeting $A$ at $(0,y,1)$ does not meet $B$ at all. This explains why $\phi$ is undefined in such cases. Correspondingly, pick any point on $B$ with $z = 0$ and any line through the origin and that point is wholly within the $xz$-plane, so will never hit $x = 1$, so is not the projection of any point on $A$, so $\phi^{-1}$ is undefined.
To understand how lines on $A$ work, think of lines as the intersection of planes. More specifically, for each line $\lambda$ in $A$ there is a unique plane $C$ through the origin such that $\lambda$ is the intersection of $A$ with $C$. Then the image under $\phi$ must be the intersection of $C$ with $B$ (since any "projection ray" from the origin through $\lambda$ lies in the plane $C$). Now, this intersection will be a line in $B$ (assuming the line was not $\{x = 0, z = 1\}$, in which case there is no intersection). So lines project to lines. Once we have that fact, it's easy to compute which line it is: just project any two points of $\lambda$, and join them up. If you really need an explicit formula, just ask.
Circles are a little trickier. Substitute $x=1/z\prime$ and $y=y\prime/z\prime$ into the equation, and get: \[\frac{1}{z^2}(y^2 + (1-z)^2)=c\]. What does this actually mean? Well, let's rearrange a little: \[\begin{align}
\frac{1}{z^2}(y^2 + 1 - 2z + z^2) &= c \\
y^2 + 1 - 2z + z^2 &= cz^2 \\
y^2 - 2z + (1-c)z^2 &= -1
\end{align}\]. At this point I want to divide by $1-c$ to complete the square, so I'm going to have to distinguish the $c=1$ case. In that case, we get \[\frac{1}{2}(y^2 + 1)=z\], which is a parabola. Otherwise: \[\begin{align}
y^2 + (1-c)(z^2 - \textstyle{\frac{2}{1-c}}z) &= -1 \\
y^2 + (1-c)((z-\textstyle{\frac{1}{1-c}})^2 - \textstyle{\frac{1}{(1-c)^2}}) &= -1 \\
y^2 + (1-c)(z-\textstyle{\frac{1}{1-c}})^2 &= \textstyle{\frac{1}{1-c}} - 1 \\
y^2 + (1-c)(z-\textstyle{\frac{1}{1-c}})^2 &= \textstyle{\frac{c}{1-c}}
\end{align}\]. For $c < 1$, this is an ellipse, while for $c > 1$, it is a hyperbola.
