Circle $\omega$ lies inside the circle $\Omega$ and touches it internally at point $P$. Point $S$ is taken on $\omega$ and the tangent to $\omega$ is drawn through it. This tangent meets $\Omega$ at points $A$ and $B$. Let $I$ be the centre of $\omega$. Find the locus of circumcentres of triangles $AIB$.
 
1. **Define the problem and given variables:**
   - Let $\omega$ be the smaller circle with center $I$ and radius $r$.
   - Let $\Omega$ be the larger circle with center $O$ and radius $R$.
   - $\omega$ touches $\Omega$ internally at point $P$.
   - Point $S$ is on $\omega$, and the tangent to $\omega$ at $S$ intersects $\Omega$ at points $A$ and $B$.
   - We need to find the locus of the circumcenters of triangles $AIB$.

2. **Identify the key geometric properties:**
   - Since $\omega$ touches $\Omega$ internally at $P$, the distance $OP = R - r$.
   - The tangent at $S$ to $\omega$ intersects $\Omega$ at $A$ and $B$.

3. **Use the Power of a Point theorem:**
   - By the Power of a Point theorem, we have:
     \[
     SA \cdot SB = SI^2
     \]
   - This is because $S$ lies on $\omega$ and $A$ and $B$ are points where the tangent from $S$ intersects $\Omega$.

4. **Determine the orthocenter of $\triangle AIB$:**
   - The orthocenter of $\triangle AIB$ is the reflection of $I$ over $S$.
   - Let $H$ be the orthocenter of $\triangle AIB$. Then $H$ is the reflection of $I$ over $S$.

5. **Use the Euler line properties:**
   - The Euler line of $\triangle AIB$ passes through the orthocenter $H$, the centroid $G$, and the circumcenter $O'$ of $\triangle AIB$.
   - The distance from the circumcenter $O'$ to the orthocenter $H$ is twice the distance from the centroid $G$ to the circumcenter $O'$.

6. **Calculate the midpoint $M$ of $AB$:**
   - Let $M$ be the midpoint of $AB$. Then $M$ lies on the Euler line of $\triangle AIB$.
   - The distance $MO'$ from $M$ to the circumcenter $O'$ is given by:
     \[
     MO' = \frac{SA \cdot SB - SI^2}{2 \cdot SI}
     \]

7. **Simplify using the Power of a Point:**
   - By the Power of a Point theorem, $SA \cdot SB = R^2 - RS^2$.
   - Therefore, we have:
     \[
     MO' = \frac{R^2 - RS^2 - SI^2}{2 \cdot SI}
     \]

8. **Use the Pythagorean theorem:**
   - By the Pythagorean theorem, $RS^2 = RI^2 + SI^2 - (RI + SI)^2$.
   - Simplifying, we get:
     \[
     RS^2 = RI^2 - SI^2
     \]

9. **Calculate the final expression for $MO'$:**
   - Substituting $RS^2$ into the expression for $MO'$, we get:
     \[
     MO' = \frac{R^2 - RI^2}{2 \cdot SI}
     \]
   - This simplifies to:
     \[
     MO' = \frac{R + RI}{2}
     \]

10. **Conclude the locus of the circumcenters:**
    - The locus of the circumcenters of triangles $AIB$ is a circle concentric with $\Omega$ with radius $\frac{2R - r}{2}$.

The final answer is $\boxed{\text{a circle concentric with } \Omega \text{ with radius } \frac{2R - r}{2}}$.