Let P be the point on the parabola $$y^2 = 4x$$ which is at the shortest distance from the center S of the circle $$x^2 + y^2 - 4x - 16y + 64 = 0$$. Let Q be the pomt on the circle dividing the line segment SP internally. Then

A

$$SP = 2\sqrt{5}$$

B

$$SQ : QP = (\sqrt{5} + 1) : 2$$

C

the x-intercept of the normal to the parabola at P is 6

D

the slope of the tangent tothe circle at Q is $$\frac{1}{2}$$