A and B are two fixed points with AB = 6 cm. A point P moves in the plane so that its distance from A is always exactly twice its distance from B. What is the path traced by P?

Two points of the line $AB$ obviously qualify: the one inside, dividing $AB$ in the ratio 2 : 1 and so 4 cm from A and 2 cm from $B$, and the one outside beyond $B$, 12 cm from A and 6 cm from $B$. Those two are the ends of a diameter of the path, and they are 12 − 4 = 8 cm apart, so the radius is 4 cm. Algebra confirms it: with A(0, 0) and $B(6,0),P{A}^2=4P{B}^2$ becomes $x^2+y^2=4[(x-6{)}^2+y^2]$, which tidies to $(x-8{)}^2+y^2=16$ — a circle of centre (8, 0) and radius 4. This is the circle of Apollonius. The perpendicular bisector is the path only for the ratio 1 : 1; no ratio other than 1 : 1 gives a straight line at all; and 3 cm would be half of $AB$, which plays no part here.