1. Given expression

We need the maximum and minimum of $f(x)=\left(\sqrt{8x-x^2-12}-4\right)^2+(x-7)^2, \quad x\in\mathbb R.$

2. Domain of definition

Since the square root must be real, $8x-x^2-12\ge 0.$ Rewrite: $-(x^2-8x+12)\ge 0$ $x^2-8x+12\le 0$ $(x-2)(x-6)\le 0.$ Hence, $x\in[2,6].$

3. Useful substitution

Observe: $8x-x^2-12=4-(x-4)^2.$ So let $y=\sqrt{4-(x-4)^2}.$ Then $y\ge 0$ and represents the upper semicircle $(x-4)^2+y^2=4.$ Now the function becomes $f(x)=(y-4)^2+(x-7)^2.$

This is the squared distance between the point $(x,y)$ on the upper semicircle $(x-4)^2+y^2=4,\quad y\ge 0$ and the fixed point $(7,4)$.

So we need the minimum and maximum squared distance from $(7,4)$ to that semicircle.

4. Geometry setup

The semicircle is part of the circle centered at $C=(4,0)$ with radius $r=2.$ The fixed point is $P=(7,4).$ Distance from center to point $P$: $CP=\sqrt{(7-4)^2+(4-0)^2}=\sqrt{9+16}=5.$

Since $P$ lies outside the circle, for the full circle:

• minimum distance = $5-2=3$
• maximum distance = $5+2=7$

We must check whether these extreme points lie on the upper semicircle.

5. Check the relevant points

The direction from $C$ to $P$ is $(3,4)$, whose unit vector is $\left(\frac35,\frac45\right).$

• Nearest point on the circle to $P$: $C+2\left(\frac35,\frac45\right)=\left(4+\frac65,\frac85\right)=\left(\frac{26}{5},\frac85\right).$ This has positive $y$, so it lies on the upper semicircle. Thus $m=3^2=9?$ Be careful: our function itself is the squared distance, so $m=3^2=9.$

• Farthest point on the circle from $P$: $C-2\left(\frac35,\frac45\right)=\left(4-\frac65,-\frac85\right)=\left(\frac{14}{5},-\frac85\right).$ This lies on the lower semicircle, not allowed. So the maximum on the upper semicircle must occur at a boundary point of the semicircle, i.e. at $x=2 \text{ or } x=6,$ where $y=0$.

6. Evaluate at boundary points

At $x=2$: $f(2)=(0-4)^2+(2-7)^2=16+25=41.$

At $x=6$: $f(6)=(0-4)^2+(6-7)^2=16+1=17.$ So the maximum is $M=41.$

And minimum is $m=9.$

7. Compute required value

We need $M^2-m^2=41^2-9^2.$ Using difference of squares, $41^2-9^2=(41-9)(41+9)=32\cdot 50=1600.$

Therefore, $\boxed{1600}.$