Animated Solution for Mathematics - Circles: Let the maximum and minimum values of $(\sqrt{8x-x^2-12}-4{)}^2+(x-7{)}^2,x\in \mathbb{R}$ be $M$ and $m$, respectively. Then $M^2-m^2$ is equal to _________

Imagine you are standing on a coordinate plane, looking at a complex algebraic expression:

At first glance, it looks like a nightmare of radicals and squares. But as an engineer, you know that every equation is just a story waiting to be told. Let's peel back the layers.

Look closely at the structure. It screams of the distance formula. If we let $y=\sqrt{8x-x^2-12}$, the expression transforms into $E=(y-4{)}^2+(x-7{)}^2$.

This is nothing more than the square of the distance between a point $(x,y)$ and a fixed point $P(7,4)$. We aren't just doing algebra; we are measuring the distance from a point to a path.

What is this path? We have $y=\sqrt{8x-x^2-12}$. Squaring both sides gives $y^2=8x-x^2-12$.

Rearranging this, we get $x^2-8x+y^2+12=0$. By completing the square, we find:

This simplifies to $(x-4{)}^2+y^2=2^2$. This is a circle centered at $C(4,0)$ with a radius of $2$.

Because the original equation involved a square root, we must have $y\ge 0$. Therefore, we are restricted to the upper semi-circle.

We need the minimum and maximum values of $E$, which is the square of the distance $d$ from $P(7,4)$ to any point on this semi-circle. First, let's find the distance from $P$ to the center $C(4,0)$:

To find the minimum distance $d_{min}$, we look along the line segment connecting $P$ to the center $C$. The closest point on the circle is $PC-r=5-2=3$.

Thus, the minimum value of the expression is $m=(d_{min}{)}^2=3^2=9$.

For the maximum distance, we look at the endpoints of our semi-circle. These occur where $y=0$, which means $(x-4{)}^2=4$, so $x=2$ or $x=6$. Our endpoints are $A(2,0)$ and $B(6,0)$.

We calculate the squared distances from $P(7,4)$ to these points:

The maximum value $M$ is clearly $41$.

We have conquered the landscape. We found $M=41$ and $m=9$. The problem asks for $M^2-m^2$:

Look at that elegance! What started as a daunting algebraic expression resolved into a clean, perfect square. This is the beauty of JEE mathematics—when you stop fighting the symbols and start visualizing the geometry, the path to the solution clears itself.