Step 1: Use the line through the centre perpendicular to the tangent.
The tangent has direction vector $(2, -1)$, so the normal direction is $(1, 2)$. The perpendicular through $(-1, 1)$ is $(x, y) = (-1 + t, 1 + 2t)$.
Step 2: Intersect with the tangent.
Substitute in $x + 2y + 4 = 0$: $(-1 + t) + 2(1 + 2t) + 4 = 0$, so $5t + 5 = 0$ and $t = -1$.
Step 3: Get the point.
\[ (x, y) = (-1 - 1,\ 1 - 2) = (-2, -1) \]
Step 4: Check the other options.
(B), (C) and (D) do not satisfy the line equation, for instance $8 - 4 + 4 \neq 0$.
Final Answer:
Option (A).
\[ \boxed{(-2, -1)} \]