Farthest a point can go

Let the any general point of the locus be $(h, k)$ .

Distance PA is given by: $\sqrt{h^2+k^2-48}$ Distance PB is given by: $\sqrt{h^2+k^2-9h-12k-12}$

$\because PA = 2PB$ :

$\sqrt{h^2+k^2-48} = 2\sqrt{h^2+k^2-9h-12k-12}$

$\implies h^2+k^2-48 = 4(h^2+k^2-9h-12k-12)$

$\implies h^2 + k^2 - 12h - 16k = 0$

To get the equation of the locus, replace $(h, k)$ with $(x, y)$ .

Therefore, required locus is:

$x^2 + y^2 - 12x - 16y = 0$

$(x-6)^2 + (y-8)^2 = 10^2$

Required locus is a circle with radius $10$ .

As the circle passes through the origin, farthest point will the point diametrically opposite, at a distance of $\boxed{20}$ .