A graph has $ n$ vertices and $ \frac {1}{2}\left(n^2 \minus{} 3n \plus{} 4\right)$ edges. There is an edge such that, after removing it, the graph becomes unconnected. Find the greatest possible length $ k$ of a circuit in such a graph.
1. **Assume the graph \( G \) is connected.**
   - Given the number of edges in the graph is \( \frac{1}{2}(n^2 - 3n + 4) \).

2. **Check the condition for the graph to have a cycle:**
   - For a connected graph with \( n \) vertices, the minimum number of edges required to ensure connectivity is \( n-1 \).
   - Therefore, we need:
     \[
     \frac{1}{2}(n^2 - 3n + 4) > n - 1
     \]
   - Simplify the inequality:
     \[
     n^2 - 3n + 4 > 2n - 2
     \]
     \[
     n^2 - 5n + 6 > 0
     \]
     \[
     (n-2)(n-3) > 0
     \]
   - This inequality holds for \( n > 3 \). Therefore, if \( G \) has a cycle, then \( n \geq 4 \).

3. **Consider the removal of an edge that disconnects the graph:**
   - Suppose removing an edge from \( G \) results in a disconnected graph \( G' \) with connected components \( G_1 \) and \( G_2 \).
   - Let \( |G_1| = n_1 \) and \( |G_2| = n_2 \) with \( n_1 + n_2 = n \).

4. **Calculate the number of edges in \( G' \):**
   - The number of edges in \( G' \) is:
     \[
     \frac{1}{2}(n^2 - 3n + 4) - 1 = \frac{1}{2}(n^2 - 3n + 2)
     \]
   - The number of edges in \( G' \) must be less than or equal to the sum of the edges in \( G_1 \) and \( G_2 \):
     \[
     \frac{1}{2}(n_1(n_1 - 1) + n_2(n_2 - 1))
     \]

5. **Set up the inequality:**
   - We have:
     \[
     \frac{1}{2}(n^2 - 3n + 2) \leq \frac{1}{2}(n_1(n_1 - 1) + n_2(n_2 - 1))
     \]
   - Simplify:
     \[
     n^2 - 3n + 2 \leq n_1^2 - n_1 + n_2^2 - n_2
     \]
   - Substitute \( n_2 = n - n_1 \):
     \[
     n^2 - 3n + 2 \leq n_1^2 - n_1 + (n - n_1)^2 - (n - n_1)
     \]
     \[
     n^2 - 3n + 2 \leq n_1^2 - n_1 + n^2 - 2nn_1 + n_1^2 - n + n_1
     \]
     \[
     n^2 - 3n + 2 \leq 2n_1^2 - 2nn_1 + n^2 - n
     \]
     \[
     -3n + 2 \leq 2n_1^2 - 2nn_1 - n
     \]
     \[
     2n_1^2 - 2nn_1 + 2n - 2 \geq 0
     \]
     \[
     2(n_1^2 - nn_1 + n - 1) \geq 0
     \]

6. **Solve the inequality:**
   - The inequality \( n_1(n_2 - 1) \leq 0 \) implies that either \( n_1 = 1 \) or \( n_2 = 1 \).
   - Without loss of generality, assume \( n_2 = 1 \). This means there is a vertex in \( G \) with degree exactly 1.

7. **Determine the maximum length \( k \) of a circuit:**
   - If there is a vertex with degree 1, the longest possible cycle in \( G \) can have at most \( n-1 \) vertices.
   - Therefore, the greatest possible length \( k \) of a circuit in such a graph is \( n-1 \).

The final answer is \( \boxed{n-1} \).