Rays $l$ and $m$ forming an angle of $a$ are drawn from the same point. Let $P$ be a fixed point on $l$. For each circle $C$ tangent to $l$ at $P$ and intersecting $m$ at $Q$ and $R$, let $T$ be the intersection point of the bisector of angle $QPR$ with $C$. Describe the locus of $T$ and justify your answer.
1. **Define the setup and initial conditions:**
   - Let rays \( l \) and \( m \) form an angle \( a \) and originate from the same point \( O \).
   - Let \( P \) be a fixed point on ray \( l \).
   - Consider a circle \( C \) tangent to \( l \) at \( P \) and intersecting \( m \) at points \( Q \) and \( R \).
   - Let \( T \) be the intersection point of the bisector of angle \( QPR \) with \( C \).

2. **Analyze the angles and properties of the circle:**
   - Let \( \angle OPQ = \theta \).
   - Since \( l \) is tangent to \( C \) at \( P \), the radius \( OP \) is perpendicular to \( l \) at \( P \).
   - Let \( P' \) be a point on \( m \) such that \( |OP| = |OP'| \). This makes \( \triangle OPP' \) isosceles with \( OP = OP' \).

3. **Determine the angles involving \( P \), \( Q \), and \( R \):**
   - Since \( \angle OPQ = \theta \), and \( \angle PRQ = \angle OPQ = \theta \) (because \( l \) is tangent to \( C \) at \( P \)).
   - Therefore, \( \angle RQP = \angle QOP + \angle OPQ = \theta + a \).

4. **Analyze the isosceles triangle \( OPP' \):**
   - Since \( \triangle OPP' \) is isosceles, \( \angle OP'P = \angle OPP' = 90^\circ - \frac{1}{2} a \).
   - Hence, \( \angle QPP' = 90^\circ - \frac{1}{2} a - \theta \).

5. **Determine the angle \( \angle P'PR \):**
   - Since \( \angle OPP' = 90^\circ - \frac{1}{2} a \) and \( \angle PP'R = 90^\circ - \frac{1}{2} a \), we have:
     \[
     \angle P'PR = 180^\circ - \angle PP'R - \angle PRP' = 180^\circ - (90^\circ + \frac{1}{2} a) - \theta = 90^\circ - \frac{1}{2} a - \theta = \angle QPP'
     \]

6. **Conclude the position of \( T \):**
   - Since \( \angle P'PR = \angle QPP' \), point \( P' \) lies on the angle bisector of \( \angle QPR \).
   - Since \( T \) also lies on the angle bisector of \( \angle QPR \), it follows that \( P \), \( P' \), and \( T \) are collinear.
   - Therefore, \( T \) lies on the line segment \( PP' \).

7. **Describe the locus of \( T \):**
   - The locus of \( T \) is the ray \( \overrightarrow{PP'} \) excluding the segment \( PP' \) itself.

The final answer is the ray \( \boxed{ \overrightarrow{PP'} } \) excluding the segment \( PP' \) itself.