Given a fixed circle $C$ and a line L through the center $O$ of $C$. Take a variable point $P$ on $L$ and let $K$ be the circle with center $P$ through $O$. Let $T$ be the point where a common tangent to $C$ and $K$ meets $K$. What is the locus of $T$?
1. **Define the problem setup:**
   - Let \( C \) be a fixed circle with center \( O \).
   - Let \( L \) be a line passing through \( O \).
   - Let \( P \) be a variable point on \( L \).
   - Let \( K \) be the circle with center \( P \) and radius \( OP \).
   - Let \( T \) be the point where a common tangent to \( C \) and \( K \) meets \( K \).

2. **Case 1: \( P \) is inside \( C \):**
   - Let \( L \) intersect \( C \) at point \( B \).
   - Let the common tangent meet \( C \) at \( Y \).
   - Let \( Z \) be the point on \( OY \) such that \( ZO = PO \).
   - Since \( ZO = PO \), \( Z \) is the reflection of \( P \) across \( O \).
   - Therefore, \( ZY = PB \) and \( ZT = TB \).
   - The angles \( \angle YZT = \angle YOP = \angle TPB \).
   - Thus, \( \triangle YZT \cong \triangle BPT \) by the Angle-Side-Angle (ASA) criterion.
   - Hence, point \( T \) lies on the perpendicular to \( L \) at \( B \).

3. **Case 2: \( P \) is outside \( C \):**
   - Let \( Z \) be the point on \( PT \) such that \( ZP = OP \).
   - Since \( ZP = OP \), \( Z \) is the reflection of \( O \) across \( P \).
   - The angles \( \angle YZT = \angle YOP = \angle TPB \).
   - Thus, \( \triangle YZT \cong \triangle TPB \) by the Angle-Side-Angle (ASA) criterion.
   - Hence, point \( T \) lies on the perpendicular to \( L \) at \( B \).

4. **Conclusion:**
   - In both cases, the point \( T \) lies on the perpendicular to \( L \) at the points of intersection of \( L \) and \( C \).

The final answer is the locus of \( \boxed{ T } \) is the perpendiculars to \( L \) at the points of intersection of \( L \) and \( C \).