a, b, x, and y = 0 is a solution, though uninteresting.

Other than that, I can only come up with the restrictions:

0 ≤ y² ≤ x² ≤ a²

0 ≤ b² ≤ x² ≤ a²

By using the fact that squares have to be positive, so x² and y² must be positive, and thus b² has to be less than or equal to x², since y² is at least 0, and at most, x².