Fig. 2
Fig. 3
Fig. 4

Solution:
a) Let $L$ and $M$ be the midpoints of the sides $CA$ and $AB$, respectively. If $O$ lies on the segment $KM$ (Fig. 2), then the segment $KM$ and the perpendicular bisector of $AB$ have two different common points $O$ and $M$, hence $KM$ is the perpendicular bisector of $AB$. Since $KM$ is parallel to $AC$ and is perpendicular to $AB$, the angle at vertex $A$ must be right. If $O$ lies on the segment $LM$ (Fig. 3), then we get similarly that the angle at vertex $B$ must be right. If $O$ lies on the segment $KL$ (Fig. 4), then on one hand $\angle ABC$ is acute, because $OK$ and $MB$ are perpendicular to $MO$, the perpendicular bisector of $AB$, and $|OK| < |LK| = |MB|$. On the other hand, $\angle ABK$ must be obtuse, since the circumcenter $O$ of the triangle $ABK$ lies outside of the triangle, a contradiction. Thus this case is not possible.

b) If the triangle $ABC$ is equilateral, then the median $AK$ is also the altitude and $ABK$ is a right triangle with the hypotenuse $AB$. The circumcenter $O$ of the last triangle is the midpoint of $AB$, i.e. an endpoint of a midsegment of the triangle, but $ABC$ is not a right triangle.