The PoincareMinkowski Map
 The map m is given by
m(x,y) = (2x/(1x^{2}y^{2}),2y/(1x^{2}y^{2}),
(1+x^{2}+y^{2})/(1x^{2}y^{2})).
 The map m maps an hline to the
intersection of H with a plane through O.
 The inverse is given by
m^{1}(x,y,(1+x^{2}+y^{2})^{½}) =
(x/(1+(1+x^{2}+y^{2})^{½}),
y/(1+(1+x^{2}+y^{2})^{½}).
Proof
(1) A point (x,y) on D corresponds to P(x,y,0) in threedimensional space.
A point on PZ' has the form t(x,y,0)+(1t)(0,0,1) = (tx,ty,t1) for some real t.
This meets H where (tx)^{2}+(ty)^{2}(t1)^{2} = 1,
i.e. 2t = t^{2}(1x^{2}y^{2}).
The root t=0 corresponds to Z' (which lies on the other sheet of the hyperboloid).
The root t=2/(1x^{2}y^{2}) gives the required point m(x,y) of H.
(3) A point P(x,y,z) lies on H has z^{2} = 1+x^{2}+y^{2}.
Since z > 0, z = (1+x^{2}+y^{2})^{½}.
Each point on PZ' has the form t(x,y,z)+(1t)(0,0,1) = (tx,ty,tz1+t).
This meets the xyplane where t=1/(1+z), i.e. at (x/(1+z),y/(1+z),0),
which corresponds to the point (x/(1+z),y/(1+z)) of D. Note that we have
z = (1+x^{2}+y^{2})^{½} > x and y, so that each point of H
corresponds to
a point of D.
(2) Suppose that the plane Π : ax+by+cz=0 actually cuts H.
By(3), these are the images of a nonempty set of points on D.
By (1), m(x,y) lies on Π if and only if
2ax/(1x^{2}y^{2})
+2by/(1x^{2}y^{2})
+c(1+x^{2}+y^{2})/(1x^{2}y^{2}) = 0.
i.e. 2ax+2by+c(1+x^{2}+y^{2}) = 0.
If c = 0 this is the line ax+by = 0, so is an hline.
If c ≠ 0, the locus has equation x^{2}+y^{2}+2(a/c)x+2(b/c)y+1 = 0.
The locus is nonempty, so (a/c)^{2}+(b/c)^{2} > 1, so, by the
Lemma
The circle C : x^{2}+y^{2}2ax2by+c = 0 is orthogonal to the circle
C : x^{2}+y^{2} = 1
if and only if c = 1 and a^{2}+b^{2} > 1.
the locus is the intersection of D with a circle orthogonal to C  an hline.

