Hydroxyketones can also form hemiacetals but, as you should expect, they usually do so less readily than hydroxy aldehydes. But we know that this hydroxyketone must exist as the cyclic hemiacetal as it has no C=O stretch in its IR spectrum. The reason? The hydroxyketone is already cyclic, with the OH group poised to attack the ketone—it can’t get away so cyclization is highly favoured.