Let $P,\ Q$ be, another pair of points on $AB,\ AC,$ as the problem states. That is, we have $BP = CQ$ and because of $BX = CY,$ we conclude that $\frac{BX}{XP} = \frac{CY}{YQ}$ $,(1)$ We denote as $O,\ K,\ L,$ the circucenters of the triangles $\triangle ABC,\ \triangle AXY,\ \triangle APQ$ respectively and let $D,\ E,\ F$ be, the ortogonal projections of $O,\ K,\ L$ respectively on $AB.$ Also, let $D',\ E',\ F'$ be, the ortogonal projections of $O,\ K,\ L$ respectively on $AC.$ Because of $D,\ E,\ F,$ are the midpoints of the segments $AB,\ AX,\ AP$ respectively, we conclude that $\frac{DE}{EF} = \frac{BX}{XP}$ $,(2)$ as well. Similarly, we have that $\frac{D'E'}{E'F'} = \frac{CY}{YQ}$ $,(3)$ From $(1),\ (2),\ (3)$ $\Longrightarrow$ $\frac{DE}{EF} = \frac{D'E'}{E'F'}$ $,(4)$ From $(4)$ we conclude that the points $O,\ K,\ L,$ are collinear, because of their ortogonal projections on two lines intersect each other, form segments with equal ratios. This is a well known and useful theorem ( easy to prove it, applying the Thales theorem ), coming from the past. In Hellenic publication is known as Θεώρημα των αναλόγων αποστάσεων $($ = Theorem of the analogous distances $)$. Hence, the locus of the circumcenter of the triangle $\triangle AXY$ as the problem states, is a line passes through the circumcenter of the given triangle $\triangle ABC$ and the problem is solved. Kostas Vittas.

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As an interesting result, we can say that the line of the wanted locus is parallel to the $A$-angle bisector of $\triangle ABC$, because of the condition of $BX = CY.$ But, I don't remember where I have posted the proof ( I don't remember it, this moment ) of the above result. So, I will search for the reference because of I am too lazy to try again to prove it. Kostas Vittas.

Quote: Let $ABC$ be a triangle and let $X\in (AB)$ , $Y\in (AC)$ be two points so that $BX = CY$ . Find the locus of the circumcenter for $\triangle XAY$ . Proof. Denote the circumcenters $O$ , $L$ of the triangles $ABC$ , $AXY$ respectively and $BX=CY=x$ . Construct the projections $O_1$ , $L_1$ of $O$ , $L$ on $AB$ respectively, the projections $O_2$ , $L_2$ of $O$ , $L$ on $AC$ respectively and the projections $U$ , $V$ of $L$ on $OO_1$ , $OO_2$ respectively. Prove easily that $O_1L_1=O_2L_2=LU=LV=\frac x2$ and $LUOV$ is a cyclical deltoid with the axis of symmetry $OL$ . Since $LU\parallel AB$ , $LV\parallel AC$ and $LO$ is bisector of $\widehat{ULV}$ obtain that $LO$ is parallelly with the $A$-bisector of $\triangle ABC$ . In conclusion, the locus of $L$ is included in a parallel line to the $A$-angle bisector of $\triangle ABC$ . Remark. Prove easily that the locus of the midpoint of $[XY]$ is included in a parrallel line to the $A$-angle bisector of $\triangle ABC$ . See another proof and other properties in the CXXIX-message from my blog.