Excellent question?

Question

If A & B are two fixed points and P is a variable point such that PA/PB is a constant (not equal to 1), then prove that P always lies on a circle intersecting AB in P & Q where P & Q divides AB internally and externally respectively in ration K:1

Answer 1

Okay. Let us define z as the distance to a point C from A opposite from B, where AB = 1, and x = AD/AB where D is a point on segment AB:

z = x^2 / (1-2x) = CA

z+1 = (x-1)^2 / (1-2x) = CB

Let us define r as the length of CP:

r = x(1-x) / (1-2x)

Then using the law of cosines, we have:

PA^2 = r^2 + z^2 - 2 r z Cos@

PB^2 = r^2 + (z+1)^2 - 2 r (z+1) Cos@

Thus, since we know that PB/PA = (1-x) / x, we ask if:

(r^2 + (z+1)^2 - 2 r (z+1) Cos@) / (r^2 + z^2 - 2 r z Cos@) = ((1-x) / x)^2

which is shown to be true for any value Cos@, proving the theorem. There isn't enough room here to show all the steps involved. As it was guessed, the loci is a circle of radius r, at distance z from A.