Given a circle with center $O$ and a point $P$ not lying on it, let $X$ be an arbitrary point on this circle and $Y$ be a common point of the bisector of angle $POX$ and the perpendicular bisector to segment $PX$.  Find the locus of points $Y$.
1. **Identify the given elements and their relationships:**
   - Let \( O \) be the center of the circle.
   - Let \( P \) be a point not on the circle.
   - Let \( X \) be an arbitrary point on the circle.
   - Let \( Y \) be the intersection of the angle bisector of \( \angle POX \) and the perpendicular bisector of segment \( PX \).

2. **Determine the properties of the angle bisector and the perpendicular bisector:**
   - The angle bisector of \( \angle POX \) divides the angle into two equal parts.
   - The perpendicular bisector of \( PX \) is the line that is perpendicular to \( PX \) at its midpoint.

3. **Consider the intersection of \( OP \) with the circle:**
   - Let \( OP \) intersect the circle at point \( Z \).
   - Since \( OX = OZ = R \) (where \( R \) is the radius of the circle), \( Z \) lies on the circle.

4. **Analyze the properties of point \( Y \):**
   - Since \( Y \) lies on the angle bisector of \( \angle POX \), it is equidistant from \( P \) and \( X \).
   - Since \( Y \) also lies on the perpendicular bisector of \( PX \), it is equidistant from \( P \) and \( X \).

5. **Combine the properties to find the locus of \( Y \):**
   - From the perpendicular bisector property, \( Y \) is equidistant from \( P \) and \( X \), so \( YP = YX \).
   - From the angle bisector property, \( Y \) is equidistant from \( P \) and \( X \), so \( YP = YX \).

6. **Conclude the locus of \( Y \):**
   - Since \( Y \) is equidistant from \( P \) and \( Z \) (because \( Y \) lies on the perpendicular bisector of \( PZ \)), the locus of \( Y \) is the perpendicular bisector of segment \( PZ \).

The final answer is the perpendicular bisector of segment \( \boxed{ PZ } \).