Prove that the orthocenter $H$ of $\triangle ABC$ belongs to their radical axis.
Denote the circles by $s_1$ and $s_2$. Let the line $A_1A_2$ intersect circles $s_1$ and $s_2$ second time in points $X_1$ and $X_2$ respectively, and the line $B_1B_2$ intersect the circles second time in points $Y_1$ and $Y_2$.
The lines $A_1Y_1$ and $A_2Y_2$ are parallel (because both of them are orthogonal to $B_1B_2$), analogously $B_1X_1$ and $B_2X_2$ are parallel. Hence these four lines form a parallelogram $KLMN$ (see fig.). It is clear that perpendiculars from the point $A$ to the line $BC$ and from the point $B$ to the line $AC$ lay on the midlines of this parallelogram. Therefore $H$ is the center of parallelogram $KLMN$ and coincide with the midpoint of segment $KM$.
In order to prove that $H$ lies on the radical axis of $s_1$ and $s_2$ it is sufficient to show that both points $K$ and $M$ belong to that radical axis.
The points $X_1$ and $Y_2$ lie on the circle $s_3$ with diameter $B_1A_2$. The line $B_1X_1$ is radical axis of $s_1$ and $s_3$, and the line $A_2Y_2$ is radical axis of $s_2$ and $s_3$. Therefore $k$ is radical center of these three circles and hence $K$ lies on the radical axis of $s_1$ and $s_2$. Analogously $M$ lies on the radical axis of $s_1$ and $s_2$.