1. Define the given parameters and setup the problem:
- Let $ABC$ be a triangle with incenter $I$.
- Given side lengths: $AB = 1400$, $AC = 1800$, $BC = 2014$.
- The circle centered at $I$ passing through $A$ intersects line $BC$ at points $X$ and $Y$.
- We need to compute the length $XY$.

2. **Calculate the semi-perimeter $s$ of $\triangle ABC$:**
$$s = \frac{AB + AC + BC}{2} = \frac{1400 + 1800 + 2014}{2} = 2607$$

3. **Determine the length $AD$ where $D$ is the foot of the altitude from $I$ to $AB$:**
$$AD = s - a = 2607 - 2014 = 593$$

4. **Express the radius $R$ of the circle centered at $I$ passing through $A$:**
$$R^2 = AD^2 + r^2 = 593^2 + r^2$$

5. **Calculate the power of point $C$ with respect to the circle $\omega$:**
- Let $E$ be the foot of the altitude from $I$ to $BC$.
- The power of $C$ is given by:
$$CI^2 - R^2 = CX \cdot CY = CY(CY + XY)$$
- We know:
$$CI^2 = CE^2 + r^2 = 1207^2 + r^2$$
- Therefore:
$$1207^2 + r^2 - (593^2 + r^2) = 1207^2 - 593^2$$

6. Simplify the expression:
$$1207^2 - 593^2 = (1207 - 593)(1207 + 593) = 614 \cdot 1800$$

7. **Calculate $CY$:**
- Since $CY = CE - YZ = 1207 - YZ = 1207 - \sqrt{R^2 - r^2} = 1207 - 593 = 614$.

8. **Determine $XY$:**
$$614 \cdot (614 + XY) = 614 \cdot 1800$$
$$614 + XY = 1800$$
$$XY = 1800 - 614 = 1186$$

The final answer is $\boxed{1186}$