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Let $N$ be the midpoint of $AH$ (and of $k_1$), and let $X$ be the image of $H$ with respect to reflection about $M$. Then $X$ lies on the circumcircle of $ABC$, opposite to $A$. As $OM$ and $AH$ are parallel, by the Intercept Theorem, we have $AH=2OM$. Hence, $AN=OM$, i.e., $ANMO$ is a parallelogram. Let $r_1$ and $r_2$ be the radii of $k_1$ and $k_2$, respectively, and let $R$ be the radius of $ABC$'s circumcircle. Then $R-r_2=OM=AN=r_1$ and, hence, $r_1+r_2=R=AO=NM$. This means that the distance between the midpoints of $k_1$ and $k_2$ is the sum of their radii. Consequently, $k_1$ and $k_2$ touch each other.