Two circles with centers $A$ and $B$ intersect at points $X$ and $Y$. The minor arc $\angle{XY}=120$ degrees with respect to circle $A$, and $\angle{XY}=60$ degrees with respect to circle $B$. If $XY=2$, find the area shared by the two circles.
1. **Understanding the problem**: We have two intersecting circles with centers \( A \) and \( B \). The minor arc \( \angle{XY} \) is \( 120^\circ \) with respect to circle \( A \) and \( 60^\circ \) with respect to circle \( B \). The length of the chord \( XY \) is given as 2 units. We need to find the area shared by the two circles.

2. **Angles and Orthogonality**: 
   - Given \( \angle AXY = 120^\circ \) and \( \angle BXY = 60^\circ \), we can deduce that \( \angle AXY + \angle BXY = 120^\circ + 60^\circ = 180^\circ \). This implies that the circles intersect orthogonally (i.e., at right angles).

3. **Radius Calculation**:
   - For circle \( A \), the central angle \( \angle XAY = 120^\circ \). Using the chord length formula in a circle, we have:
     \[
     XY = 2R \sin\left(\frac{\theta}{2}\right)
     \]
     where \( R \) is the radius and \( \theta \) is the central angle. For circle \( A \):
     \[
     2 = 2R \sin\left(\frac{120^\circ}{2}\right) = 2R \sin(60^\circ) = 2R \cdot \frac{\sqrt{3}}{2} = R\sqrt{3}
     \]
     Thus, \( R = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \).

   - For circle \( B \), the central angle \( \angle XBY = 60^\circ \). Using the same chord length formula:
     \[
     2 = 2r \sin\left(\frac{60^\circ}{2}\right) = 2r \sin(30^\circ) = 2r \cdot \frac{1}{2} = r
     \]
     Thus, \( r = 2 \).

4. **Area Calculation**:
   - The area of the sector \( \angle XAY \) in circle \( A \):
     \[
     \text{Area of sector } XAY = \frac{120^\circ}{360^\circ} \cdot \pi R^2 = \frac{1}{3} \cdot \pi \left(\frac{2\sqrt{3}}{3}\right)^2 = \frac{1}{3} \cdot \pi \cdot \frac{4 \cdot 3}{9} = \frac{4\pi}{9}
     \]

   - The area of the sector \( \angle XBY \) in circle \( B \):
     \[
     \text{Area of sector } XBY = \frac{60^\circ}{360^\circ} \cdot \pi r^2 = \frac{1}{6} \cdot \pi \cdot 2^2 = \frac{1}{6} \cdot \pi \cdot 4 = \frac{2\pi}{3}
     \]

5. **Area of Quadrilateral \( AXBY \)**:
   - The area of quadrilateral \( AXBY \) can be found by summing the areas of the two sectors and subtracting the area of the shared region \( x \):
     \[
     \text{Area of } AXBY = \text{Area of sector } XAY + \text{Area of sector } XBY - x
     \]
     Given that the circles are orthogonal, the area of quadrilateral \( AXBY \) can be calculated using the formula for the area of a cyclic quadrilateral:
     \[
     \text{Area of } AXBY = \frac{1}{2} \cdot XY \cdot AB = \frac{1}{2} \cdot 2 \cdot \sqrt{R^2 + r^2} = \sqrt{3} + \frac{\sqrt{3}}{3} = \frac{4\sqrt{3}}{3}
     \]

6. **Solving for \( x \)**:
   \[
   \frac{4\pi}{9} + \frac{2\pi}{3} - x = \frac{4\sqrt{3}}{3}
   \]
   Simplifying the left-hand side:
   \[
   \frac{4\pi}{9} + \frac{6\pi}{9} - x = \frac{4\sqrt{3}}{3}
   \]
   \[
   \frac{10\pi}{9} - x = \frac{4\sqrt{3}}{3}
   \]
   Solving for \( x \):
   \[
   x = \frac{10\pi}{9} - \frac{4\sqrt{3}}{3}
   \]

The final answer is \( \boxed{\frac{10\pi - 12\sqrt{3}}{9}} \).