If $ (x^2+1)(y^2+1)+9=6(x+y)$ where $x,y$ are real numbers, what is $x^2+y^2$?

$ 
\textbf{(A)}\ 7
\qquad\textbf{(B)}\ 6 
\qquad\textbf{(C)}\ 5
\qquad\textbf{(D)}\ 4
\qquad\textbf{(E)}\ 3
$
1. Given the equation:
   \[
   (x^2+1)(y^2+1) + 9 = 6(x+y)
   \]
   We aim to find \(x^2 + y^2\).

2. By the Cauchy-Schwarz Inequality, we have:
   \[
   (x^2+1)(y^2+1) \geq (xy + 1)^2
   \]
   However, a more straightforward approach is to use the AM-GM inequality:
   \[
   (x^2+1)(y^2+1) \geq (x+y)^2
   \]
   This implies:
   \[
   (x^2+1)(y^2+1) \geq (x+y)^2
   \]
   Substituting this into the given equation:
   \[
   (x+y)^2 + 9 \leq 6(x+y)
   \]

3. Let \(a = x + y\). Then the inequality becomes:
   \[
   a^2 + 9 \leq 6a
   \]
   Rearranging terms, we get:
   \[
   a^2 - 6a + 9 \leq 0
   \]
   This can be factored as:
   \[
   (a-3)^2 \leq 0
   \]
   Since the square of a real number is non-negative, the only solution is:
   \[
   a - 3 = 0 \implies a = 3
   \]
   Therefore, \(x + y = 3\).

4. Substituting \(x + y = 3\) back into the original equation:
   \[
   (x^2+1)(y^2+1) + 9 = 6 \cdot 3
   \]
   Simplifying, we get:
   \[
   (x^2+1)(y^2+1) + 9 = 18
   \]
   \[
   (x^2+1)(y^2+1) = 9
   \]

5. Expanding the left-hand side:
   \[
   x^2y^2 + x^2 + y^2 + 1 = 9
   \]
   Since \(x + y = 3\), we can use the identity \((x+y)^2 = x^2 + y^2 + 2xy\):
   \[
   3^2 = x^2 + y^2 + 2xy
   \]
   \[
   9 = x^2 + y^2 + 2xy
   \]

6. Let \(s = x^2 + y^2\) and \(p = xy\). Then:
   \[
   s + 2p = 9
   \]
   Also, from the expanded equation:
   \[
   p^2 + s + 1 = 9
   \]
   \[
   p^2 + s = 8
   \]

7. Solving the system of equations:
   \[
   s + 2p = 9
   \]
   \[
   p^2 + s = 8
   \]
   Subtract the second equation from the first:
   \[
   (s + 2p) - (p^2 + s) = 9 - 8
   \]
   \[
   2p - p^2 = 1
   \]
   Rearrange to form a quadratic equation:
   \[
   p^2 - 2p + 1 = 0
   \]
   \[
   (p-1)^2 = 0
   \]
   \[
   p = 1
   \]

8. Substitute \(p = 1\) back into \(s + 2p = 9\):
   \[
   s + 2 \cdot 1 = 9
   \]
   \[
   s + 2 = 9
   \]
   \[
   s = 7
   \]

The final answer is \(\boxed{7}\).