Plane Geometry - Bruce E. Shapiro

298 SECTION 51. THE POINCARE DISK MODELFigure 51.1: Orthogonal circles. The tangents (or radii) of each circle meetat a 90 ◦ angle at the points of intersection.O=(0,0)r=1aSubstituting equation 51.2 into equation 51.1 givesx 2 − 2xx 0 + y 2 − 2yy 0 + 1 = 0 (51.3)Suppose now that we want the circle to pass through the pointsP = (x 1 , y 1 ), Q = (x 2 , y 2 )This gives two equations in the unknowns x 0 and y 0 :x 2 1 − 2x 1 x 0 + y1 2 − 2y 1 y 0 + 1 = 0x 2 2 − 2x 2 x 0 + y2 2 − 2y 2 y 0 + 1 = 0Rearranging, Å ã Å ãx1 y 1 2x0x 2 y 2 2y 0Å ã 1 + r2= 11 + r22where r i = x 2 i + y2 i , for i = 1, 2. HenceÅ ã 2x0= 1 Å ã Å ãy2 −y 1 1 + r212y 0 ∆ −x 2 x 1 1 + r22where ∆ is the determinant(51.4)(51.5)∆ = x 1 y 2 − x 2 y 1« CC BY-NC-ND 3.0. Revised: 18 Nov 2012