In a triangle $$PQR$$, the co-ordinates of the points $$P$$ and $$Q$$ are $$(-2, 4)$$ and $$(4, -2)$$ respectively. If the equation of the perpendicular bisector of $$PR$$ is $$2x - y + 2 = 0$$, then the centre of the circumcircle of the $$\triangle PQR$$ is:

We have triangle $$PQR$$ with $$P(-2, 4)$$ and $$Q(4, -2)$$. The perpendicular bisector of $$PR$$ is $$2x - y + 2 = 0$$. We need the circumcentre.

The circumcentre lies on the perpendicular bisector of every side of the triangle.

The midpoint of $$PQ$$ is $$\left(\frac{-2+4}{2}, \frac{4+(-2)}{2}\right) = (1, 1)$$.

The slope of $$PQ$$ is $$\frac{-2-4}{4-(-2)} = \frac{-6}{6} = -1$$. So the perpendicular bisector of $$PQ$$ has slope $$1$$ and passes through $$(1, 1)$$: $$y - 1 = 1(x - 1)$$, i.e., $$y = x$$ $$-(1)$$.

The perpendicular bisector of $$PR$$ is given as $$2x - y + 2 = 0$$ $$-(2)$$.

The circumcentre lies on both $$(1)$$ and $$(2)$$. From $$(1)$$: $$y = x$$. Substituting into $$(2)$$: $$2x - x + 2 = 0$$, so $$x = -2$$ and $$y = -2$$.

The circumcentre is $$(-2, -2)$$, which matches Option B.