The inverse in a circle
Author: Chris Sangwin
Date: 23 Dec 2006

In this worksheet the point P is free to move on the circle through A and B.  The point P' is on the line OP.  The distance from O is 1/OP.  Hence the position of P' is often described as the "inverse in a circle" of P.  Move A, B and then move P around the circle.  Can you find a configuration in which P' moves along an exact straight line?

O = (0, 0)
r = 1
c = Circle(O, r)
A = (0.6950617283950618, 0.38641975308641974)
B = (1.1592592592592592, 0.2876543209876543)
d = Circle(A, B)
P = Point(d)
b = Ray(O, P)
r': 1 / Distance[O, P]
e = Circle(O, r')
P' = Intersect(e, b, 1)