http://mathworld.wolfram.com/FirstEppsteinPoint.html http://mathworld.wolfram.com/SecondEppsteinPoint.html

does somebody have a solution?

my solution was based on inversion, but there are other solutions for example using homothety.....

We invert about the tangency point of two of the circles, to obtain an equivalent problem. Let $l_1, l_2$ be two parallel lines. Circles $C_1$ is tangent to $l_1,l_2$ at $W,X$ respectively, and circle $C_2$ is tangent to $l_1,l_2,C_1$ at $Y,Z,T$, respectively. Let $O$ be the centre of inversion. Prove that the circumcircles of $\triangle WOZ, \triangle YOX$ intersect at a point distinct from $O$ lying on $OT$. Clearly, $O$ lies on the radical axis of both circumcircles. As $W,T,Z$ and $Y,T,X$ are collinear, and $WT \cdot TZ = YT \cdot TX$, $T$ lies on the radical axis of both circumcircles. Thus $OT$ is the radical axis of the circumcircles, so they have an intersection point other than $O$ with $OT$.