1. Find the center of the circle

The circle is $x^2+y^2-4x-16y+64=0.$

Complete squares: $x^2-4x+y^2-16y+64=0$ $(x-2)^2-4+(y-8)^2-64+64=0$ $(x-2)^2+(y-8)^2=4.$

So the center is $S=(2,8),$ and radius is $r=2.$

2. Point on parabola nearest to $S$

The parabola is $y^2=4x.$ A standard parametric point on it is $P=(t^2,2t).$

Distance squared from $S(2,8)$ to $P$ is $SP^2=(t^2-2)^2+(2t-8)^2.$ Expand: $SP^2=t^4-4t^2+4+4t^2-32t+64=t^4-32t+68.$

To minimize distance, minimize $f(t)=t^4-32t+68.$

Differentiate: $f'(t)=4t^3-32=4(t^3-8).$ Set to zero: $t^3=8 \implies t=2.$

Also, $f''(t)=12t^2>0$ at $t=2$, so this gives minimum.

Hence $P=(2^2,2\cdot 2)=(4,4).$

Now $SP=\sqrt{(4-2)^2+(4-8)^2}=\sqrt{4+16}=\sqrt{20}=2\sqrt5.$

So Option A is correct.

3. Find point $Q$ on the circle dividing $SP$ internally

Since $Q$ lies on the circle and on segment $SP$, it is the point where segment $SP$ meets the circle.

Because $S$ is the center, along the line from $S$ to $P$, the point on the circle is at distance equal to radius from $S$. Thus, $SQ=r=2.$ Also, $SP=2\sqrt5.$ So $QP=SP-SQ=2\sqrt5-2=2(\sqrt5-1).$

Therefore, $SQ:QP=2:2(\sqrt5-1)=1:(\sqrt5-1).$

Rationalize/comparison: $\frac{1}{\sqrt5-1}=\frac{\sqrt5+1}{4},$ so $1:(\sqrt5-1)=(\sqrt5+1):4.$

But the option says $SQ:QP=(\sqrt5+1):2,$ which is not equal.

So Option B is incorrect.

4. Normal to parabola at $P$ and its $x$-intercept

For parabola $y^2=4x$, differentiate: $2y\frac{dy}{dx}=4$ $\frac{dy}{dx}=\frac{2}{y}.$

At $P=(4,4)$, $m_{\text{tangent}}=\frac{2}{4}=\frac12.$ So slope of normal is $m_{\text{normal}}=-2.$

Equation of normal through $(4,4)$: $y-4=-2(x-4).$ So $y=-2x+12.$

Its $x$-intercept is found by putting $y=0$: $0=-2x+12 \implies x=6.$

So Option C is correct.

5. Slope of tangent to the circle at $Q$

First find line $SP$.

Points are $S=(2,8)$ and $P=(4,4)$, so slope of $SP$ is $m_{SP}=\frac{4-8}{4-2}=\frac{-4}{2}=-2.$

Since $Q$ lies on line $SP$, radius $SQ$ has slope $-2$. The tangent to the circle at $Q$ is perpendicular to radius $SQ$. Hence its slope is the negative reciprocal of $-2$: $m_{\text{tangent at }Q}=\frac12.$

So Option D is correct.

6. Final selection

Correct options are:

• A
• C
• D

This matches the stored correct answer.