Given a circle with fixed points $A$ and $B$ on its circumference, and $XY$ as a variable diameter of the circle, we are to determine the locus of the point of intersection of lines $AX$ and $BY$. We assume that $AB$ is not a diameter of the circle.

### Step-by-step Solution:

1. Understanding the Problem:

- Let $O$ be the center of the circle.
- The line $XY$ is a variable diameter, which means $O$ is the midpoint of $XY$.
- The lines $AX$ and $BY$ are drawn such that $X$ and $Y$ can vary along the circumference due to the diameter condition.

2. Geometric Analysis:

- Since $XY$ is a diameter, the angle $\angle XOY = 180^\circ$.
- According to the properties of a circle, any angle subtended by a diameter on the circle is a right angle. Thus, both $\angle XAY = 90^\circ$ and $\angle XBY = 90^\circ$ when $X$ and $Y$ lie on the same circle.

3. Finding the Locus:

- Consider the triangle $\triangle AXB$. The point of intersection of lines $AX$ and $BY$, denoted as $P$, must satisfy certain constraints due to the varying diameter.

- Since $XY$ is a diameter, any such $P$ forms two pairs of right angles with the ends of the diameter: $\angle XAY = \angle XBY = 90^\circ$.

- This observation implies that point $P$ lies on the circle known as the \textbf{nine-point circle} (or Feuerbach circle) of triangle $\triangle AOB$.

- However, since both angles remain consistent as $X$ and $Y$ traverse the circle, the locus traced by $P$ indeed forms another circle, as the configuration is symmetric with respect to the circle's center and varies consistently irrespective of specific arcs.

Thus, the locus of the points of intersections of lines $AX$ and $BY$ as $XY$ runs over all possible diameters is a circle. Therefore, the final answer is:
$$\boxed{\text{a circle}}$$