Let $C$ be a circle with radius $R$ and centre $O$, and $S$ a fixed point in the interior of $C$.  Let $AA'$ and $BB'$ be perpendicular chords through $S$. Consider the rectangles $SAMB$, $SBN'A'$, $SA'M'B'$, and $SB'NA$.  Find the set of all points $M$, $N'$, $M'$, and $N$ when $A$ moves around the whole circle.
1. **Understanding the Problem:**
   - We have a circle \( C \) with radius \( R \) and center \( O \).
   - \( S \) is a fixed point inside the circle.
   - \( AA' \) and \( BB' \) are perpendicular chords passing through \( S \).
   - We need to find the set of all points \( M \), \( N' \), \( M' \), and \( N \) when \( A \) moves around the circle.

2. **Analyzing the Geometry:**
   - The quadrilateral \( ABA'B' \) is cyclic with perpendicular diagonals, implying that its anticenter (the concurrency point of its maltitudes) is identical with its diagonal intersection \( S \).
   - Let \( P, Q, P', Q' \) be the midpoints of the sides \( AB, BA', A'B', B'A \) respectively.
   - The centroid \( G \) of the cyclic quadrilateral \( ABA'B' \) (the intersection of lines \( PP', QQ' \) connecting the midpoints of the opposite sides) is the midpoint between its circumcenter \( O \) and anticenter \( S \).

3. **Properties of the Parallelogram:**
   - Since \( PQP'Q' \) is a parallelogram, its diagonals \( PP', QQ' \) cut each other in half at \( G \).
   - Thus, the diagonals \( OS, PP' \) of the quadrilateral \( OPSP' \) also cut each other in half, making this quadrilateral a parallelogram.

4. **Using the Parallelogram Rule:**
   \[
   PP'^2 + OS^2 = 2(OP^2 + SP^2)
   \]
   \[
   GP^2 + GS^2 = \frac{OP^2 + SP^2}{2}
   \]

5. **Calculating Distances:**
   - The segment \( SP \) is the median of the right-angle triangle \( \triangle ABS \), hence \( AP = BP = SP \).
   - Since \( P \) is the midpoint of the chord \( AB \) of the circle \( O \), \( OP \perp AB \) and:
     \[
     OP^2 + SP^2 = OP^2 + AP^2 = OA^2 = R^2
     \]
   - Consequently, the distance \( GP \) is given by:
     \[
     GP^2 = \frac{R^2}{2} - GS^2
     \]
   - Similarly:
     \[
     GP'^2 = GQ^2 = GQ'^2 = \frac{R^2}{2} - GS^2
     \]

6. **Locus of Points:**
   - The midpoints \( P, P', Q, Q' \) all lie on the circle with center \( G \) and radius independent of the direction of the perpendicular lines \( AA' \perp BB \) meeting at \( S \).
   - The distance \( MM' \) between the vertices \( M, M' \) of the rectangles \( SABM, SA'B'M' \) is twice the distance \( PP' \) between their diagonal intersections \( P, P' \).
   - Since the midpoint of \( PP' \) is \( G \), the midpoint of \( MM' \) is on the line \( GS \) twice the distance \( GS \) from the common vertex \( S \) of the two rectangles, i.e., the midpoint of \( MM' \) is \( O \), the center of the circle \( O \).

7. **Conclusion:**
   - The locus of \( M, M' \) (and similarly, the locus of \( N, N' \)) is a circle concentric with the circle \( O \) and with radius:
     \[
     r = 2 GP = 2 \sqrt{\frac{R^2}{2} - GS^2} = \sqrt{2R^2 - OS^2}
     \]

The final answer is \( \boxed{ \sqrt{2R^2 - OS^2} } \).