The function $f$, with domain on the set of non-negative integers, is defined by the following :
$\bullet$ $f (0) = 2$
$\bullet$ $(f (n + 1) -1)^2 + (f (n)-1) ^2 = 2f (n) f (n + 1) + 4$, taking $f (n)$ the largest possible value.
Determine $f (n)$.
1. **Simplify the given equation:**
   The given functional equation is:
   \[
   (f(n+1) - 1)^2 + (f(n) - 1)^2 = 2f(n)f(n+1) + 4
   \]
   Let \( u = f(n+1) \) and \( v = f(n) \). Substituting these into the equation, we get:
   \[
   (u - 1)^2 + (v - 1)^2 = 2uv + 4
   \]
   Expanding and simplifying:
   \[
   u^2 - 2u + 1 + v^2 - 2v + 1 = 2uv + 4
   \]
   \[
   u^2 + v^2 - 2u - 2v + 2 = 2uv + 4
   \]
   \[
   u^2 + v^2 - 2uv - 2u - 2v - 2 = 0
   \]
   \[
   u^2 + (-2v - 2)u + v^2 - 2v - 2 = 0
   \]

2. **Solve the quadratic equation for \( u \):**
   The quadratic equation in \( u \) is:
   \[
   u^2 + (-2v - 2)u + v^2 - 2v - 2 = 0
   \]
   Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -2v - 2 \), and \( c = v^2 - 2v - 2 \):
   \[
   u = \frac{2v + 2 \pm \sqrt{(2v + 2)^2 - 4(v^2 - 2v - 2)}}{2}
   \]
   \[
   u = \frac{2v + 2 \pm \sqrt{4v^2 + 8v + 4 - 4v^2 + 8 + 8}}{2}
   \]
   \[
   u = \frac{2v + 2 \pm \sqrt{16v + 20}}{2}
   \]
   \[
   u = \frac{2v + 2 \pm 2\sqrt{4v + 5}}{2}
   \]
   \[
   u = v + 1 \pm \sqrt{4v + 5}
   \]
   Since \( f(n) \) is the largest possible value, we take the positive root:
   \[
   f(n+1) = v + 1 + \sqrt{4v + 5}
   \]
   \[
   f(n+1) = f(n) + 1 + \sqrt{4f(n) + 5}
   \]

3. **Conjecture and prove by induction:**
   We conjecture that \( f(n) = n^2 + n\sqrt{11} + 2 \). We will prove this by induction.

   **Base case:**
   For \( n = 0 \):
   \[
   f(0) = 2
   \]
   which matches the given \( f(0) = 2 \).

   **Inductive step:**
   Assume \( f(k) = k^2 + k\sqrt{11} + 2 \) for some \( k \in \mathbb{N} \). We need to show that \( f(k+1) = (k+1)^2 + (k+1)\sqrt{11} + 2 \).

   Using the recursive formula:
   \[
   f(k+1) = f(k) + 1 + \sqrt{4f(k) + 5}
   \]
   Substitute \( f(k) = k^2 + k\sqrt{11} + 2 \):
   \[
   f(k+1) = k^2 + k\sqrt{11} + 2 + 1 + \sqrt{4(k^2 + k\sqrt{11} + 2) + 5}
   \]
   \[
   f(k+1) = k^2 + k\sqrt{11} + 3 + \sqrt{4k^2 + 4k\sqrt{11} + 8 + 5}
   \]
   \[
   f(k+1) = k^2 + k\sqrt{11} + 3 + \sqrt{4k^2 + 4k\sqrt{11} + 13}
   \]
   \[
   f(k+1) = k^2 + k\sqrt{11} + 3 + \sqrt{(2k + \sqrt{11})^2}
   \]
   \[
   f(k+1) = k^2 + k\sqrt{11} + 3 + 2k + \sqrt{11}
   \]
   \[
   f(k+1) = k^2 + 2k + 1 + 2 + (k + 1)\sqrt{11}
   \]
   \[
   f(k+1) = (k+1)^2 + (k+1)\sqrt{11} + 2
   \]
   This completes the inductive step.

   Therefore, by mathematical induction, \( f(n) = n^2 + n\sqrt{11} + 2 \) for all \( n \in \mathbb{N} \).

The final answer is \( \boxed{ f(n) = n^2 + n\sqrt{11} + 2 } \)