How to visualize a 2D cartesian point in homogeneous coordinates? Geometric primitives
Hi, I am trying to understand below text from a book on Computer Vision by Szeliski.

I am not able to visualize it combining cartesian and homogeneous coordinates. Below is my attempt in understanding visually. Kindly correct and confirm.
Further, if correct, what are A and B as shown in circle? Afaik,
$$
B = \sqrt{(\tilde x)^2 + (\tilde y)^2 + (\tilde w)^2} = 1? 
$$
because, w=1 there?

 A: One way to look at it is that each point in the projective plane corresponds to a line through the origin of the three-dimensional Cartesian space,
and each such line corresponds to a point in the projective plane.
If you set $B = \sqrt{\tilde x^2 + \tilde y^2 + \tilde w^2} = 1$ 
then you will only reach points on the unit sphere around the origin of the three-dimensional Cartesian space. Every line through the origin intersects that sphere,
so you'll capture all the lines (and all the projective points) that way,
but except for $(0,0,1),$ none of the points on that sphere is actually on the plane $\tilde w = 1.$
If you want to plot $(x,y)$ at $(\tilde x, \tilde y, 1)$ as in the figure,
then at each such point you have $A = \tilde w = 1$
and $B = \sqrt{\tilde x^2 + \tilde y^2 + \tilde w^2} = 
 \sqrt{x^2 + y^2 + 1}.$ 
I'm not sure what to do with those facts, however.
If you then take any point $(\tilde x, \tilde y, \tilde w)$ 
on any of the lines through the origin in the three-dimensional space,
assuming you don't pick the origin $(0,0,0)$ itself, then there are just two cases to think about:
Case $\tilde w = 0$: the line is in the $\tilde x, \tilde y$ plane and does not intersect the plane $\tilde w$ at all.
So this line corresponds to one of the "points at infinity" that you can adjoin to the ordinary Cartesian plane in order to make it a projective plane.
Case $\tilde w \neq 0$: the line intersects the plane $\tilde w$ at the point
$$(x,y,1) = \left(\frac{\tilde x}{\tilde w}, \frac{\tilde x}{\tilde w}, \frac{\tilde w}{\tilde w}\right).$$
But something to remember is that if you're using a line through the origin to represent a projective point, any point on that line (other than the origin itself) is as good as any other point on that line.
