I feel that using fractional ought to help somehow. Consider a circle C_a through 0 with center at the point a in C. What does it become upon application of 1/z ? We know fractional linear transformations take lines and circles to lines and circles. 1/C_a "passes through oo", so must be a straight line. Which one? The point 2a is the farthest point from 0 of C_a, so it follows that 1/(2a) is the farthest point from oo of 1/C_a, i.e. the nearest to 0. Thus 1/C_a is the line L_a passing through 1/(2a) perpendicular to 1/(2a), i.e., L_a(t) = 1/(2a) + t a*, t in R , where z* denotes conj(z). Now let a and b denote z_1 and z_2, respectively. (*) We exclude the cases Ka + Lb = 0 for some K, L in R as exceptional. Applying 1/z to both circles C_a and C_b, we get the lines L_a(s) = 1/(2a) + s a*, s in R and L_b(t) = 1/(2b) + t b*, t in R whose intersection is at, say, (s,t) = (S,T). At this stage I don't know a painless way to get the complex intersection point of L_a and L_b. But when you find it, say at the point J(a,b), the original second intersection will be at its reciprocal, 1/J(a,b). Condition (*) above suggests there should be a denominator of the form a*b - ab* — or someting like that! — in the expression, since that being 0 is equivalent (maybe after adjustment) to (*). —Dan
On Jan 28, 2016, at 2:14 PM, Bill Gosper <billgosper@gmail.com> wrote:
. . . For the Clifford circle theorems I needed second intersection: Given centers of circles though the origin z1, z2,
secondintersection[z1_, z2_] := (Conjugate[z1] z2 - Conjugate[z2] z1)/Conjugate[z1 - z2]
Is this obvious to anybody? I had to derive it with 8th grade analytic geometry.