Let $x$ and $y$ be real numbers satisfying the equation $x^2-4x+y^2+3=0$. If the maximum and minimum values of $x^2+y^2$ are $M$ and $m$ respectively, compute the numerical value of $M-m$.
1. Start with the given equation:
   \[
   x^2 - 4x + y^2 + 3 = 0
   \]

2. Rearrange the equation to complete the square for \(x\):
   \[
   x^2 - 4x + 4 + y^2 + 3 - 4 = 0
   \]
   \[
   (x - 2)^2 + y^2 - 1 = 0
   \]
   \[
   (x - 2)^2 + y^2 = 1
   \]
   This represents a circle with center \((2, 0)\) and radius \(1\).

3. We need to find the maximum and minimum values of \(x^2 + y^2\). This is the square of the distance from any point \((x, y)\) on the circle to the origin \((0, 0)\).

4. The distance \(d\) from the origin to the center of the circle \((2, 0)\) is:
   \[
   d = \sqrt{(2 - 0)^2 + (0 - 0)^2} = \sqrt{4} = 2
   \]

5. The maximum distance from the origin to a point on the circle is the sum of the distance from the origin to the center and the radius of the circle:
   \[
   d_{\text{max}} = 2 + 1 = 3
   \]
   Therefore, the maximum value of \(x^2 + y^2\) is:
   \[
   M = d_{\text{max}}^2 = 3^2 = 9
   \]

6. The minimum distance from the origin to a point on the circle is the difference between the distance from the origin to the center and the radius of the circle:
   \[
   d_{\text{min}} = 2 - 1 = 1
   \]
   Therefore, the minimum value of \(x^2 + y^2\) is:
   \[
   m = d_{\text{min}}^2 = 1^2 = 1
   \]

7. The difference between the maximum and minimum values of \(x^2 + y^2\) is:
   \[
   M - m = 9 - 1 = 8
   \]

The final answer is \(\boxed{8}\).