For the real pairs $(x,y)$ satisfying the equation $x^2 + y^2 + 2x - 6y = 6$, which of the following cannot be equal to $(x-1)^2 + (y-2)^2$?

$ 
\textbf{(A)}\ 2
\qquad\textbf{(B)}\ 9
\qquad\textbf{(C)}\ 16
\qquad\textbf{(D)}\ 23
\qquad\textbf{(E)}\ 30
$
1. Start with the given equation:
   \[
   x^2 + y^2 + 2x - 6y = 6
   \]
   We can complete the square for both \(x\) and \(y\).

2. For \(x\):
   \[
   x^2 + 2x = (x+1)^2 - 1
   \]

3. For \(y\):
   \[
   y^2 - 6y = (y-3)^2 - 9
   \]

4. Substitute these into the original equation:
   \[
   (x+1)^2 - 1 + (y-3)^2 - 9 = 6
   \]

5. Simplify the equation:
   \[
   (x+1)^2 + (y-3)^2 - 10 = 6
   \]
   \[
   (x+1)^2 + (y-3)^2 = 16
   \]
   This represents a circle centered at \((-1, 3)\) with radius 4.

6. We need to determine the possible values of \((x-1)^2 + (y-2)^2\). This represents the square of the distance from \((x, y)\) to \((1, 2)\).

7. The distance from the center \((-1, 3)\) to the point \((1, 2)\) is:
   \[
   \sqrt{(1 - (-1))^2 + (2 - 3)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}
   \]

8. Using the triangle inequality, the distance from any point \((x, y)\) on the circle to \((1, 2)\) must be between:
   \[
   4 - \sqrt{5} \quad \text{and} \quad 4 + \sqrt{5}
   \]

9. Squaring these distances:
   \[
   (4 - \sqrt{5})^2 = 16 - 8\sqrt{5} + 5 = 21 - 8\sqrt{5}
   \]
   \[
   (4 + \sqrt{5})^2 = 16 + 8\sqrt{5} + 5 = 21 + 8\sqrt{5}
   \]

10. Therefore, the possible values of \((x-1)^2 + (y-2)^2\) lie in the interval:
    \[
    21 - 8\sqrt{5} \quad \text{to} \quad 21 + 8\sqrt{5}
    \]

11. Calculate the approximate values:
    \[
    21 - 8\sqrt{5} \approx 21 - 8 \times 2.236 \approx 21 - 17.888 \approx 3.112
    \]
    \[
    21 + 8\sqrt{5} \approx 21 + 8 \times 2.236 \approx 21 + 17.888 \approx 38.888
    \]

12. From the given options, the value 2 is not within the interval \([3.112, 38.888]\).

The final answer is \(\boxed{2}\)