Given triangle $\triangle ABC$. Consider an Apollonius circle $\omega_1$ of segment $AB$ with ratio $BC:CA$, i.e. for all points $X$ of this circle, we obtain: $\frac{XA}{XB}=\frac{BC}{CA}$. We define the Apollonius circle $\omega_2$ for segment $BC$ analogously - for all points $Y\in\omega_2$, we obtain: $\frac{YB}{YC}=\frac{CA}{AB}$. Prove, that if $\omega_1$ and $\omega_2$ do intersect, then both of the points of intersection lay on the Euler line of triangle $\triangle ABC$, in other words, that the Euler line of $\triangle ABC$ is the radical axis of $\omega_1$ and $\omega_2$.
I don't know how to make use of the Euler line at all and would be really, really grateful for any hints or solutions :)