97. Let $P$ be the foot of the perpendicular dropped from $N$ to the line $M B$; then $|M P'| = R \cos \alpha$, hence $|M P|$ is equal to the distance from the center $O$ of $A B$, but the distance from the vertex of the triangle to the orthocenter is twice the distance from the center of the circumscribed circle to the opposite side, i.e., $|M P| = \frac{1}{2} |M K|$.

From this it follows that if $M$ is on the larger arc, i.e., $\widehat{A M B} = \alpha$, then $|M K| = R$; if, however, $\widehat{A M B} = 180^{\circ} - \alpha$ (i.e., $M$ is on the smaller arc of the circle), then $|N K|^{2} = R^{2} + 4 R^{2} \cos^{2} \alpha + 4 R^{2} \cos^{2} \alpha = R^{2} \left(1 + 8 \cos^{2} \alpha\right)$.

Answer: $|M K| = \left\{\begin{array}{l}R, \text { if } M \text { is on the larger arc of the circle, } \\ R \sqrt{1 + 8 \cos^{2} \alpha}, \\ \text { if } M \text { is on the smaller arc of the circle. }\end{array}\right.$