Solution:
If $AB$ is a diameter of $k$, then $AN$ and $BM$ are altitudes of $\triangle ABC$. Let $H$ be the orthocenter of $\triangle ABC$ and let the tangent line to $k$ at $M$ meet the altitude $CH$ at $O_1$. Then $\text{CMO 1 = ABM = AM 2}$ and $\angle ABM = \angle ACH$. Thus $\angle CMO_1 = \angle MCO_1$, i.e. $CO_1 = MO_1$. On the other hand $\angle O_1HM = 90^\circ - \angle O_1CM = 90^\circ - \angle CMO_1 = \angle O_1MH$, i.e. $O_1M = O_1H$. Therefore $O_1$ is the midpoint of $CH$. It can be seen analogously that the tangent line to $k$ at $N$ passes through $O_1$, i.e. $O \equiv O_1$ and $OM = ON = OC = CH/2$.

Let $O$ be the circumcenter of $\triangle CMN$. Then $\angle CMO = \angle MCO = \angle ABM$ and $\angle CNO = \angle NCO = \angle BAN$. Hence
$$\angle ACB = \angle MCO + \angle NCO = \angle ABM + \angle BAN.$$
Therefore
$$\begin{aligned} 2\angle ANB &= \angle ANB + \angle AMB \\ &= 180^\circ - \angle ABC - \angle BAN + 180^\circ - \angle BAC - \angle ABM \\ &= 360^\circ - (\angle ABC + \angle ACB + \angle BAC) = 180^\circ. \end{aligned}$$
Hence $\angle ANB = 90^\circ$ and $AB$ is a diameter of $k$.