Let $O$ be the centre of a circle and $A$ a fixed interior point of the circle different from $O$.  Determine all points $P$ on the circumference of the circle such that the angle $OPA$ is a maximum.

[asy]
import graph;

unitsize(2 cm);

pair A, O, P;

A = (0.5,0.2);
O = (0,0);
P = dir(80);

draw(Circle(O,1));
draw(O--A--P--cycle);

label("$A$", A, E);
label("$O$", O, S);
label("$P$", P, N);
[/asy]
1. **Define the problem geometrically**:
   - Let \( O \) be the center of the circle.
   - Let \( A \) be a fixed interior point of the circle different from \( O \).
   - We need to determine all points \( P \) on the circumference of the circle such that the angle \( \angle OPA \) is maximized.

2. **Consider the circle and the given points**:
   - Let the radius of the circle be \( R \).
   - Let \( O \) be the origin of the coordinate system, i.e., \( O = (0,0) \).
   - Let \( A \) be a fixed point inside the circle, i.e., \( A = (a,b) \) where \( \sqrt{a^2 + b^2} < R \).

3. **Analyze the angle \( \angle OPA \)**:
   - The angle \( \angle OPA \) is formed by the vectors \( \overrightarrow{OP} \) and \( \overrightarrow{PA} \).
   - To maximize \( \angle OPA \), we need to consider the geometric properties of the circle and the points involved.

4. **Use the property of the circle**:
   - The angle \( \angle OPA \) is maximized when \( P \) is such that the line segment \( OP \) is perpendicular to the line segment \( PA \).
   - This is because the maximum angle in a triangle is achieved when one of the sides is perpendicular to the other side.

5. **Determine the locus of points \( P \)**:
   - For \( \angle OPA \) to be maximized, \( P \) must lie on the circle such that \( OP \perp PA \).
   - This implies that \( P \) must be the point where the perpendicular from \( A \) to the line \( OP \) intersects the circle.

6. **Find the perpendicular bisector**:
   - The perpendicular bisector of \( OA \) will pass through the midpoint of \( OA \) and be perpendicular to \( OA \).
   - Let \( M \) be the midpoint of \( OA \). Then \( M = \left( \frac{a}{2}, \frac{b}{2} \right) \).

7. **Equation of the perpendicular bisector**:
   - The slope of \( OA \) is \( \frac{b}{a} \).
   - The slope of the perpendicular bisector is \( -\frac{a}{b} \).
   - The equation of the perpendicular bisector is:
     \[
     y - \frac{b}{2} = -\frac{a}{b} \left( x - \frac{a}{2} \right)
     \]
     Simplifying, we get:
     \[
     y = -\frac{a}{b} x + \frac{a^2 + b^2}{2b}
     \]

8. **Intersection with the circle**:
   - The circle has the equation \( x^2 + y^2 = R^2 \).
   - Substitute \( y = -\frac{a}{b} x + \frac{a^2 + b^2}{2b} \) into the circle's equation:
     \[
     x^2 + \left( -\frac{a}{b} x + \frac{a^2 + b^2}{2b} \right)^2 = R^2
     \]
   - Solve this equation to find the points of intersection, which will give the points \( P \) on the circumference.

9. **Solve the quadratic equation**:
   - After substituting and simplifying, we get a quadratic equation in \( x \).
   - Solve for \( x \) and then find the corresponding \( y \) values.

10. **Conclusion**:
    - The points \( P \) that maximize \( \angle OPA \) are the points of intersection of the perpendicular bisector of \( OA \) with the circle.

The final answer is the points of intersection of the perpendicular bisector of \( \boxed{ OA } \) with the circle.