Inside a circle of center $ O$ and radius $ r$, take two points $ A$ and $ B$ symmetrical about $ O$. We consider a variable point $ P$ on the circle and draw the chord $ \overline{PP'}\perp \overline{AP}$. Let $ C$ is the symmetric of $ B$ about $ \overline{PP'}$  ($ \overline{PP}'$ is the axis of symmetry) . Find the locus of point $ Q \equal{} \overline{PP'}\cap\overline{AC}$ when we change $ P$ in the circle.
1. **Symmetry and Reflection:**
   - Given a circle with center \( O \) and radius \( r \).
   - Points \( A \) and \( B \) are symmetrical about \( O \), meaning \( O \) is the midpoint of \( \overline{AB} \).
   - Point \( P \) is a variable point on the circle.
   - Chord \( \overline{PP'} \) is perpendicular to \( \overline{AP} \).
   - Point \( C \) is the reflection of \( B \) across \( \overline{PP'} \).

2. **Reflection of \( A \):**
   - Let \( D \) be the reflection of \( A \) across \( \overline{PP'} \).
   - Since \( \overline{PP'} \) is the perpendicular bisector of both \( \overline{AC} \) and \( \overline{BD} \), \( ABCD \) forms an isosceles trapezoid with \( \overline{PP'} \) as the axis of symmetry.

3. **Intersection Point \( Q \):**
   - The point \( Q \) is defined as the intersection \( \overline{PP'} \cap \overline{AC} \).
   - Since \( \overline{PP'} \) is the perpendicular bisector of \( \overline{AC} \), \( Q \) lies on \( \overline{PP'} \).

4. **Isosceles Triangle Property:**
   - In \( \triangle QCB \), since \( C \) is the reflection of \( B \) across \( \overline{PP'} \), \( \triangle QCB \) is isosceles with \( QB = QC \).

5. **Constant Sum of Distances:**
   - Since \( \overline{AC} = 2r \) (as \( A \) and \( C \) are on the circle and \( \overline{AC} \) is a diameter), we have:
     \[
     QA + QC = AC = 2r
     \]
   - This implies that the sum of the distances from \( Q \) to \( A \) and \( Q \) to \( C \) is constant.

6. **Locus of \( Q \):**
   - The locus of point \( Q \) is an ellipse with foci at \( A \) and \( B \) and the major axis length \( 2r \).
   - The line \( \overline{PP'} \) is always tangent to this ellipse.

The final answer is the locus of \( \boxed{ Q } \) is an ellipse with foci \( A \) and \( B \) and major axis length \( 2r \).