projectivization of a hyperplane? I was studying An Introduction to Hyperplane
Arrangements by
Richard P. Stanley that I ran into this on page 5:
"This provides a good way to visualize three-dimensional real
linear arrangements. For instance, if [arrangement] $A$ consists of the three coordinate hyperplanes
$x_1 = 0, x_2 = 0,$ and$ x_3 = 0$ , then a projective drawing is given by

The line labelled $i$ is the projectivization of the hyperplane $x_i = 0$. The hyperplane
at infinity is $x_3 = 0$. There are four regions, so$ r(A) = 8$. To draw the incidences
among all eight regions of $A$, simply reflect the interior of the disk to the exterior:

"
how is projectivization of the hyperplane $x_i = 0$ is a circle?
 A: I don't know what "This provides..." in your quote is. You should add some content from the book preceding the excerpt you have quoted. But extrapolating from your pictures, assume you have a finite plane arrangement with at least three planes in the projective three space. Every three planes determine a common point of intersection. You end up with a finite set of such intersection points. Draw a (projective) plane that does not go through any of these intersection points formed by triples of planes from the arrangement. The intersection of the planes with the fixed new plane is an arrangement of lines in that plane. Think of the projective three space as the Euclidean three space augmented with a plane at infinity. Then place the unit round sphere on top of the Euclidean plane and for each point on the sphere, draw the line passing through that point and the center of the sphere. The latter line intersects the plane in a unique point defining a map from the sphere to the plane. Notice that diametrically opposite point to the point on the sphere maps to the same point on the plane so the map is two to one. Moreover, each great circle on the sphere maps to a straight line on the plane and each line on the plane is the image of a unique great circle on the sphere. The equator of the sphere, i.e. the great circle parallel to the plane, is mapped to the line at infinity. The affine part of the plane is bijectively mapped to the lower hemisphere of the sphere (below the equator) and the lane at infinity is mapped to the equator but two to one. You end up with an arrangement of great circles on the round sphere. Now if you pick the north pole, i.e. the point diametrically opposite to the point of tangency of the sphere and the plane, and perform stereographic projection from the north pole onto the plane, the lower open (southern) hemisphere is mapped inside the open disc of radius two centered at the point of tangency between the sphere and the plane. The circular boundary of the disc is the image of the line at infinity of the plane, but mapped in a two to one way. Under the stereographic projection each great circle on the sphere is mapped to a circular arc in the disc with endpoints diametrically opposite on the circular boundary of the disc. Consequently, we get a one to one correspondence between lines in the plane and the circular arcs with diametrically opposite endpoints on the boundary of the disc. Consequently, the arrangement of planes in three space  becomes an arrangement of lines in the plane which in their own turn become an arrangement of circular arcs with diametrically opposite endpoints on the boundary of the disc. In particular if two lines of the plane pass through the tangency point with the sphere, they correspond to a pair of great circles through the north pole and are consequently mapped to two straight line diameters of the disc. So you can see how this picture you have appears naturally.                         
