In the plane let $\,C\,$ be a circle, $\,L\,$ a line tangent to the circle $\,C,\,$ and $\,M\,$ a point on $\,L$. Find the locus of all points $\,P\,$ with the following property: there exists two points $\,Q,R\,$ on $\,L\,$ such that $\,M\,$ is the midpoint of $\,QR\,$ and $\,C\,$ is the inscribed circle of triangle $\,PQR$.
1. **Define the elements and setup the problem:**
   - Let \( C \) be a circle with center \( O \) and radius \( r \).
   - Let \( L \) be a line tangent to \( C \) at point \( T \).
   - Let \( M \) be a point on \( L \).
   - We need to find the locus of all points \( P \) such that there exist points \( Q \) and \( R \) on \( L \) with \( M \) as the midpoint of \( QR \) and \( C \) as the incircle of triangle \( PQR \).

2. **Identify key points and their properties:**
   - Let \( T' \) be the point diametrically opposite \( T \) on \( C \).
   - Let \( S \) be the reflection of \( T \) over \( M \).

3. **Establish the collinearity condition:**
   - For \( C \) to be the incircle of \( \triangle PQR \), the point \( P \) must lie on the line \( \overline{T'S} \). This is because the homothety centered at \( P \) that maps \( C \) to the \( P \)-excircle must send \( T \) to \( S \).

4. **Determine the valid region for \( P \):**
   - \( P \) must lie on the ray starting from \( T' \) and extending in the direction of \( S \). 
   - If \( P \) is closer to \( S \) than \( T' \) or if \( P = T' \), it is impossible to draw the tangents \( \overline{PQ} \) and \( \overline{PR} \) such that \( C \) is the incircle.

5. **Verify the construction:**
   - For any point \( P \) on the ray \( \overline{T'S} \) (excluding \( T' \)), the tangents from \( P \) to \( C \) will meet \( L \) at points \( Q \) and \( R \).
   - The midpoint \( M' \) of \( \overline{QR} \) must coincide with \( M \) because \( T \) is the reflection of \( S \) over \( M \).

6. **Conclusion:**
   - The locus of all points \( P \) is the ray starting from \( T' \) and extending in the direction of \( S \), excluding \( T' \).

\(\blacksquare\)

The final answer is the ray starting from \( \boxed{ T' } \) and extending in the direction of \( S \), excluding \( T' \).