I'm reading a text (https://people.maths.ox.ac.uk/hitchin/files/LectureNotes/Projective_geometry/Chapter_2_Quadrics.pdf) to study about Quadrics in the projective space. The following is the example (1) (on Page 26).
Example
Consider the hyperbola $xy = 1$. Using the homogeneous coordinate ($x = x_1/x_0, y = x_2/x_0$), we can transform into; $$ \left( \frac{1}{2}(x_1 + x_2) \right)^2 - \left( \frac{1}{2}(x_1 - x_2) \right)^2 - x_0^2 = 0 $$
and then if (x_1 + x_2 \neq 0), we put
$$ y_1 = \frac{x_1 - x_2}{x_1 + x_2}, \quad y_2 = \frac{2x_0}{x_1 + x_2} $$
and the conic intersects the copy of $\mathbb{R}^2 \subset P^2(\mathbb{R})$ (the complement of the line $x_1 + x_2 = 0$) in the circle;$y_1^2 + y_2^2 = 1$.
The original line at infinity $x_0 = 0$ meets this in $y_2 = 0$.
Question
I understood that we transform the hyperbola ($x_1 x_2 - x^2_0 = 0$ in $P^2(\mathbb{R})$ into a circle in $\mathbb{R^2}$. But I can't quite imagine how $x_0$ and $y_2$ meet. Could someone help me understand this or give me some visual aid?
In 3-space it's a linear transformation taking one cone with singular point the origin to another, one with axis the $y=x$ in the $xy$ plane and the other the $z$-axis. However, the $x_0=1$ in one is $y_2=2$ in the other, and $y_0=1$ in the other is $x_1+x_2=1$ in the first. Only trouble is, one cone isn't right circular and the other is (i.e. the transformation is not just a rotation; there will be some stretching and contracting going on), so we get that $x_1+x_2=1$ intersects the cone $x_0^2=x_1x_2$ in a circle of radius $\frac{\sqrt3}2,$ and not $1.$
Similarly the cone $y_1^2+y_2^2=y_0^2$ intersects in $y_2=2$ to get a hyperbola, but not quite a rectangular one, like the one we started out with. 