Let a point $$P$$ be such that its distance from the point (5, 0) is thrice the distance of $$P$$ from the point (-5, 0). If the locus of the point $$P$$ is a circle of radius $$r$$, then $$4r^2$$ (in the nearest integer) is equal to ______.

Let $$P = (x, y)$$. The condition is that the distance from $$P$$ to $$(5, 0)$$ is three times the distance from $$P$$ to $$(-5, 0)$$. Squaring both sides: $$(x - 5)^2 + y^2 = 9[(x + 5)^2 + y^2]$$.

Expanding: $$x^2 - 10x + 25 + y^2 = 9x^2 + 90x + 225 + 9y^2$$. Rearranging: $$8x^2 + 100x + 8y^2 + 200 = 0$$, which simplifies to $$x^2 + \frac{25}{2}x + y^2 + 25 = 0$$.

Completing the square: $$\left(x + \frac{25}{4}\right)^2 + y^2 = \frac{625}{16} - 25 = \frac{625 - 400}{16} = \frac{225}{16}$$.

This is a circle with radius $$r = \frac{15}{4}$$, so $$r^2 = \frac{225}{16}$$ and $$4r^2 = \frac{225}{4} = 56.25$$.

Rounding to the nearest integer, $$4r^2 = 56$$.

The correct answer is $$56$$.