A line in the plane of a triangle $ABC$ intersects the sides $AB$ and $AC$ respectively at points $X$ and $Y$ such that $BX = CY$ . Find the locus of the center of the circumcircle of triangle $XAY .$
1. **Define the problem and setup:**
   Let \( \triangle ABC \) be a triangle with points \( X \) and \( Y \) on sides \( AB \) and \( AC \) respectively such that \( BX = CY = x \). We need to find the locus of the circumcenter \( L \) of \( \triangle XAY \).

2. **Introduce circumcenters and projections:**
   Denote the circumcenters of \( \triangle ABC \) and \( \triangle XAY \) as \( O \) and \( L \) respectively. Construct the projections \( O_1 \) and \( L_1 \) of \( O \) and \( L \) on \( AB \) respectively, and the projections \( O_2 \) and \( L_2 \) of \( O \) and \( L \) on \( AC \) respectively. Also, construct the projections \( U \) and \( V \) of \( L \) on \( OO_1 \) and \( OO_2 \) respectively.

3. **Establish equal distances:**
   Since \( BX = CY = x \), we have:
   \[
   O_1L_1 = O_2L_2 = LU = LV = \frac{x}{2}
   \]
   This is because the projections of the circumcenter on the sides of the triangle are equidistant from the points where the sides are intersected by the line.

4. **Cyclic quadrilateral and symmetry:**
   The quadrilateral \( LUOV \) is cyclic because the opposite angles are supplementary. The axis of symmetry of this cyclic quadrilateral is \( OL \). Since \( LU \parallel AB \) and \( LV \parallel AC \), and \( LO \) is the bisector of \( \angle ULV \), it follows that \( LO \) is parallel to the \( A \)-angle bisector of \( \triangle ABC \).

5. **Conclusion about the locus:**
   Since \( LO \) is parallel to the \( A \)-angle bisector of \( \triangle ABC \), the locus of \( L \) (the circumcenter of \( \triangle XAY \)) is a line parallel to the \( A \)-angle bisector of \( \triangle ABC \).

The final answer is the locus of the circumcenter of \( \boxed{ \triangle XAY } \) is a line parallel to the \( A \)-angle bisector of \( \triangle ABC \).