Given a circle $K$, find the locus of vertices $A$ of parallelograms $ABCD$ with diagonals $AC \leq BD$, such that $BD$ is inside $K$.
1. **Understanding the Problem:**
   We are given a circle \( K \) and need to find the locus of vertices \( A \) of parallelograms \( ABCD \) with diagonals \( AC \leq BD \), such that \( BD \) is inside \( K \).

2. **Condition Analysis:**
   The condition \( AC \leq BD \) implies that the angle \( \angle A \) in the parallelogram must be at least \( 90^\circ \). This is because in a parallelogram, the diagonals bisect each other, and if \( AC \leq BD \), the angle \( \angle A \) must be greater than or equal to \( 90^\circ \).

3. **Geometric Interpretation:**
   For \( BD \) to be inside the circle \( K \), the point \( A \) must be positioned such that the diagonals \( BD \) and \( AC \) satisfy the given conditions. If \( A \) is inside the circle \( K \), then \( \angle A \) can be \( 90^\circ \) or more, satisfying \( AC \leq BD \).

4. **Considering \( A \) Outside the Circle:**
   If \( A \) is outside the circle \( K \), the maximum possible \( \angle A \) occurs when \( AB \) and \( AD \) are tangents to the circle \( K \). The tangents from a point outside a circle form equal angles with the line joining the point to the center of the circle. Therefore, the locus of such points \( A \) forms a circle with the same center as \( K \) but with a radius \( \sqrt{2} \) times that of \( K \).

5. **Conclusion:**
   The locus of points \( A \) is the circle with the same center as \( K \) and radius \( \sqrt{2} \) times the radius of \( K \).

The final answer is the circle with the same center as \( \boxed{ K } \) and radius \( \sqrt{2} \) times the radius of \( K \).