Solution. Define $P=\overline{AE}\cap \overline{BC}$ and $Q=\overline{AC}\cap \overline{BE}$. Clearly $Q$ is the orthocenter of $\bigtriangleup APB$ and $EPCQ$ is cyclic. In this vein, it's well-known that $OC$ and $OE$ are tangent to $(EPCQ)$ (simple angle-chasing), thus $$KC^2=KQ\cdot KE$$i.e. $Q$ is the inverse of $E$ with respect to $c_1$ so it lies on its polar $ST$. The result follows. $\blacksquare$

Nice problem! My solution: It suffices to show that the poles of $ST,BE,AC$ wrt $c_1$ are collinear. Let the common tangent of $c,c_1$ at $C$ intersect $AE$ at $P$, and let $D$ be the second intersection of $AC$ with $c_1$. Then $C,E$ lie on the circle $c_2$ with diameter $KP$. Since the homothety centered at $C$ that sends $c_1$ to $c$ maps $D$ to $A$, we have $DK \parallel AB$. This yields $\angle ECD = \angle ECA = \angle EBA = \angle EKD$, i.e. $D$ also lies on $c_2$. But then $DK \perp DP$, which means that $DP$ is tangent to $c_1$. Thus $P$ is the pole of the line $\overline{ACD}$ and $E$ is the pole of $ST$ wrt $c_1$. Since $EP \perp BE$ and $BE$ passes through $K$, the pole of $BE$ is $\infty_{EP}$, and the claim follows.

Let D be the intersection of AC and BE.Since ABCE is cyclic we get that BEC and BAC are equal angles, from where we get that DCK and CEK are similar triangles.Then EK • DK = CK^2.Since ST is the polar of E with respect to c_1,we obtain that ST passes through D.

This is exactly Balkan MO 2010 G8!