A and B are two points on a co-ordinate plane. All the points, in the same plane as A and B, whose distance from B is twice that from A lie on
a a straight line intersecting AB at a point O such that 2AO = BO.
b a circle with center at a point O on AB such that AO = 2BO.
c a circle with center at a point O on AB extended such that 4AO = BO..
d None of these.
Suppose that $A(0,0)$, $B(a, 0)$ and $C(x,y)$.
$$2CA=CB \implies$$
$$2\sqrt{x^2+y^2}=\sqrt{(x-a)^2+y^2}$$
$$4(x^2+y^2)=(x-a)^2+y^2$$
$$4x^2+4y^2=x^2-2ax+a^2+y^2$$
$$3x^2+2ax+3y^2=a^2$$
$$x^2+\frac23ax+y^2=\frac13a^2$$
$$(x+\frac13a)^2-\frac19a^2+y^2=\frac13a^2$$
$$(x+\frac13a)^2+y^2=\frac49a^2$$
...which is a circle with center $O(-\frac13a,0)$.
It means that $OA=\frac13a$, $OB=\frac43a$ or $OB=4OA$.
So the right answer is answer (c).