Let $P$ be the point on the parabola $y ^ { 2 } = 4 x$ which is at the shortest distance from the center $S$ of the circle $x ^ { 2 } + y ^ { 2 } - 4 x - 16 y + 64 = 0$. Let $Q$ be the point on the circle dividing the line segment $S P$ internally. Then\\
(A) $S P = 2 \sqrt { 5 }$\\
(B) $S Q : Q P = ( \sqrt { 5 } + 1 ) : 2$\\
(C) the $x$-intercept of the normal to the parabola at $P$ is 6\\
(D) the slope of the tangent to the circle at $Q$ is $\frac { 1 } { 2 }$