Let $P$ be a point outside the circle $C$. Find two points $Q$ and $R$ on the circle, such that $P,Q$ and $R$ are collinear and $Q$ is the midpopint of the segmenet $PR$. (Discuss the number of solutions).
1. Let \( C \) be a circle with center \( O \) and radius \( R \). Let \( P \) be a point outside the circle \( C \).
2. We need to find points \( Q \) and \( R \) on the circle such that \( P, Q, \) and \( R \) are collinear and \( Q \) is the midpoint of the segment \( PR \).
3. Since \( Q \) is the midpoint of \( PR \), we have \( PQ = QR \). Let \( d = PO \) be the distance from \( P \) to the center \( O \) of the circle.
4. By the Power of a Point theorem, the power of point \( P \) with respect to the circle is given by:
   \[
   \text{Power of } P = PO^2 - R^2
   \]
5. Since \( Q \) is the midpoint of \( PR \), we have:
   \[
   PQ = QR = \frac{PR}{2}
   \]
6. Let \( PQ = x \). Then \( PR = 2x \).
7. Using the Power of a Point theorem, we have:
   \[
   PQ \cdot PR = PO^2 - R^2
   \]
   Substituting \( PR = 2x \) and \( PQ = x \), we get:
   \[
   x \cdot 2x = PO^2 - R^2
   \]
   Simplifying, we obtain:
   \[
   2x^2 = PO^2 - R^2
   \]
8. Solving for \( x \), we get:
   \[
   x^2 = \frac{PO^2 - R^2}{2}
   \]
   \[
   x = \sqrt{\frac{PO^2 - R^2}{2}}
   \]
9. Since \( x = PQ \), we have:
   \[
   PQ = \sqrt{\frac{PO^2 - R^2}{2}}
   \]
10. The number of solutions depends on the value of \( PO^2 - R^2 \). If \( PO^2 - R^2 > 0 \), then there are two possible points \( Q \) and \( R \) on the circle such that \( P, Q, \) and \( R \) are collinear and \( Q \) is the midpoint of \( PR \). If \( PO^2 - R^2 = 0 \), then there is exactly one solution. If \( PO^2 - R^2 < 0 \), then there are no real solutions.

The final answer is \( \boxed{ \leq 2 } \) solutions.