a geometry problem - Apollonius circles

A geometry problem PhL 8.18.09 Problem: what locus is described by | z - a | = α |z-b| α > 0 That is to say, the ratio of the distances from two points is a constant. Solution: Square both sides and do the usual stuff: (x-ax)2 + (y-ay)2 = α2 { (x-bx)2 + (y-by)2 } (x-ax)2 - α2(x-bx)2 + (y-ay)2 - α2 (y-by)2 = 0 (1-α2) x2 +(-2ax+2α2bx)x + (1-α2) y2 + (-2ay+2α2by)y + ( ax2- α2bx2 + ay2- α2by2 ) = 0 x2 +(-2ax+2α2bx)/ (1-α2) x + y2 + (-2ay+2α2by)/ (1-α2) y + ( ax2- α2bx2 + ay2- α2by2 )/ (1-α2) = 0 x2 +Ax'x + y2 + Ay' y + B = 0 where Ax'/2 = (-ax+α2bx)/ (1-α2) Ay'/2 = (-ay+α2by)/ (1-α2) B = ( ax2- α2bx2 + ay2- α2by2 )/ (1-α2) Now complete the square, x2 +Ax'x + (Ax'/2)2+ y2 + Ay' y + (Ay'/2)2 + B - (Ax'/2)2- (Ay'/2)2 = 0 (x + Ax'/2)2+ (y + Ay'/2)2 + (B - (Ax'/2)2- (Ay'/2)2 ) = 0 This is a circle! The center of the circle is at x0 = - Ax'/2 = - (-ax+α2bx)/ (1-α2) y0 = - Ay'/2 = - (-ay+α2by)/ (1-α2) The squared radius of the circle is this: R2 = (Ax'/2)2+ (Ay'/2)2 - B = - ( ax2- α2bx2 + ay2- α2by2 )/ (1-α2) + (-ax+α2bx)2/ (1-α2)2+ (-ay+α2by)2/ (1-α2)2 So this is the circle implied by | z - a| = α |z-b|. Now first let's put a and b on the x axis. Then x0 = - Ax'/2 = - (-ax+α2bx)/ (1-α2) y0 = - Ay'/2 = - (-ay+α2by)/ (1-α2) R2 = - ( ax2- α2bx2 + ay2- α2by2 )/ (1-α2) + (-ax+α2bx)2/ (1-α2)2+ (-ay+α2by)2/ (1-α2)2 x0 = - Ax'/2 = - (-ax+α2bx)/ (1-α2) y0 = - Ay'/2 = - (-0+α20)/ (1-α2) = 0 R2 = - ( ax2- α2bx2 + 02- α202 )/ (1-α2) + (-ax+α2bx)2/ (1-α2)2+ (-0+α20)2/ (1-α2)2 = - ( ax2- α2bx2)/ (1-α2) + (-ax+α2bx)2/ (1-α2)2 1/(1-α2)2 * { - ( ax2- α2bx2)(1-α2) + (-ax+α2bx)2 } Continuing {} = -ax2 + α2bx2 + ax2 + α4bx2 - 2α2axbx + α2(ax2- α2bx2) = α2bx2 + α4bx2 - 2α2axbx + α2(ax2- α2bx2) = α2[bx2 + α2bx2 - 2axbx + (ax2- α2bx2) ] = α2[bx2 + α2bx2 - 2axbx + ax2- α2bx2 ] = α2[bx2 - 2axbx + ax2] = α2(bx- ax)2 To summarize, in the case where a and b are on the real axis, | z - a| = α |z-b| is a circle where: x0 = - (-ax+α2bx)/ (1-α2) y0 = 0 R2 = α2/(1-α2)2 * (bx - ax)2 R = { α / |1-α2|} | bx - ax | I think this situation must look like this. Imagine α = 2 so that R = (2/3) | bx - ax | = 2. where I have done a precision scaled drawing. Notice that both left and right extremes of the circle give the right answer: 2/1 and 6/3. So the circle notion must be right. Summary: Consider the locus of points in the complex plane that solve this equation: | z - a | = α |z-b| => α > 0 A. The solution locus is a circle of some radius R and center location c. B. The circle center c is on the line connecting the points a and b, and lies outside the segment ab. C. If α > 1, then the ordering of these points is a, b, c as shown in this picture, which is correctly scaled, D. If α < 1, then the ordering is c, a, b and the circle encloses point a and not point b. This situation is obtained from C by doing the swap a↔b. E. The specific center and radius of the circle are given by: cx = - (-ax+α2bx)/ (1-α2) cy = - (-ay+α2by)/ (1-α2) R2 = cx2+ cy2 - ( ax2- α2bx2 + ay2- α2by2 )/ (1-α2) F. If we put a and b on the x-axis, these results simplify to cx = - (-ax+α2bx)/ (1-α2) cy = 0 R = { α / |1-α2|} | bx - ax | G. If in addition we put point a at the origin, we get cx = - (α2bx)/ (1-α2) cy = 0 R = { α / |1-α2|} | bx | |cx| = (α2|bx|)/ |1-α2| = αR H. If α = 1+ then R = ∞ and cx = +∞. Think if this as the limit of the above picture as α reduces from 2 to 1. The center c moves out to the right and the radius increases. The circle becomes a vertical line bisecting the segment ab, and of course this is true for arbitrary a and b.