Let be given points $A,B$ on a circle and let $P$ be a variable point on that circle. Let point $M$ be determined by $P$ as the point that is either on segment $PA$ with $AM=MP+PB$ or on segment $PB$ with $AP+MP=PB$. Find the locus of points $M$.
1. **Given Points and Circle:**
   Let \( A \) and \( B \) be fixed points on a circle, and let \( P \) be a variable point on the circle. We need to find the locus of point \( M \) such that:
   - \( M \) is on segment \( PA \) with \( AM = MP + PB \), or
   - \( M \) is on segment \( PB \) with \( AP + MP = PB \).

2. **Analyzing the Condition \( AM = MP + PB \):**
   - From \( AM = MP + PB \), we can rewrite it as \( AM - MP = PB \).
   - Let \( M \) be the point on segment \( PA \) such that \( AM = MP + PB \). This implies that \( M \) divides \( PA \) in such a way that \( AM = MP + PB \).

3. **Analyzing the Condition \( BM = MP + PA \):**
   - From \( BM = MP + PA \), we can rewrite it as \( BM - MP = PA \).
   - Let \( M \) be the point on segment \( PB \) such that \( BM = MP + PA \). This implies that \( M \) divides \( PB \) in such a way that \( BM = MP + PA \).

4. **Using Perpendicular Diameter \( CD \):**
   - Let \( CD \) be the diameter of the circle perpendicular to \( AB \).
   - Suppose \( P \) is such that \( AP > PB \). We use the condition \( AM = MP + PB \).

5. **Constructing Point \( E \):**
   - On the extension of \( AP \), construct point \( E \) such that \( PE = PB \).
   - Then, \( AM = MP + PE \Rightarrow AM = ME \Rightarrow M \) is the midpoint of \( AE \).

6. **Analyzing \( P \in \stackrel{\frown}{BD} \):**
   - We need to prove that point \( D \) is on the perpendicular bisector of \( BE \).
   - The external angle bisector of the isosceles triangle \( PBE \) is the line \( PC \). Since \( PC \perp PD \), the internal angle bisector of \( \triangle PBE \) is the line \( PD \).
   - Since \( \triangle PBE \) is isosceles, \( PD \) is also the perpendicular bisector of its base \( BE \).

7. **Perpendicular Bisectors:**
   - \( CD \) is the perpendicular bisector of \( AB \), so \( DA = DB = DE \).
   - Therefore, point \( E \) moves on the circle with center \( D \) and radius \( DA \).

8. **Locus of Point \( M \):**
   - Since \( DM \perp AM \), point \( M \) lies on the circle with diameter \( AD \).
   - Similarly, for other arcs:
     - \( P \in \stackrel{\frown}{BC} \Rightarrow M \) lies on the circle with diameter \( AC \).
     - \( P \in \stackrel{\frown}{AC} \Rightarrow M \) lies on the circle with diameter \( BC \).
     - \( P \in \stackrel{\frown}{AD} \Rightarrow M \) lies on the circle with diameter \( BD \).

The locus of points \( M \) consists of the red arcs on the circles with diameters \( AD \), \( AC \), \( BC \), and \( BD \).

The final answer is the locus of points \( \boxed{ M } \) on the red arcs of the circles with diameters \( AD \), \( AC \), \( BC \), and \( BD \).