the poincare-minkowski map

 The Poincare-Minkowski Map The map m is given by m(x,y) = (2x/(1-x2-y2),2y/(1-x2-y2), (1+x2+y2)/(1-x2-y2)). The map m maps an h-line to the intersection of H with a plane through O. The inverse is given by m-1(x,y,(1+x2+y2)½) = (x/(1+(1+x2+y2)½), y/(1+(1+x2+y2)½). Proof (1) A point (x,y) on D corresponds to P(x,y,0) in three-dimensional space. A point on PZ' has the form t(x,y,0)+(1-t)(0,0,-1) = (tx,ty,t-1) for some real t. This meets H where (tx)2+(ty)2-(t-1)2 = -1, i.e. 2t = t2(1-x2-y2). The root t=0 corresponds to Z' (which lies on the other sheet of the hyperboloid). The root t=2/(1-x2-y2) gives the required point m(x,y) of H. (3) A point P(x,y,z) lies on H has z2 = 1+x2+y2. Since z > 0, z = (1+x2+y2)½. Each point on PZ' has the form t(x,y,z)+(1-t)(0,0,-1) = (tx,ty,tz-1+t). This meets the xy-plane 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+x2+y2)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 non-empty set of points on D. By (1), m(x,y) lies on Π if and only if 2ax/(1-x2-y2) +2by/(1-x2-y2) +c(1+x2+y2)/(1-x2-y2) = 0. i.e. 2ax+2by+c(1+x2+y2) = 0. If c = 0 this is the line ax+by = 0, so is an h-line. If c ≠ 0, the locus has equation x2+y2+2(a/c)x+2(b/c)y+1 = 0. The locus is non-empty, so (a/c)2+(b/c)2 > 1, so, by the Lemma The circle C : x2+y2-2ax-2by+c = 0 is orthogonal to the circle C : x2+y2 = 1 if and only if c = 1 and a2+b2 > 1. the locus is the intersection of D with a circle orthogonal to C - an h-line.