At the triangle $ABC$ with angles $\alpha, \beta, \gamma$ the points $A'$, $B'$ and $C'$ are between $B$ and $C$, $A$ and $C$, $A$ and $B$ respectively, such that the circumcircles of $AB'C'$ and $A'BC'$ have only the point $C'$ in common.
It holds that the two circumcircles have the same tangent in $C'$ and that the angle $\angle A'C'B'$ is equal to $\alpha+\beta$.
Show that the $C'$ is on the circumcircle of $A'B'C$.
What condition must be satisfied so that $C'$ is on the circumcircle of $A'B'C$ ? Do we have to Find the radius of the circle and the calculate the distance of the center to $C'$ ?
You are almost there. Look at the $A'C'B'C$ quadrilateral. You have $\angle A'C'B'=\alpha+\beta$. In the $ABC$ triangle you have $$\angle ACB=\angle B'CA'=180^\circ-\alpha-\beta$$ Then $$\angle B'CA'+\angle A'C'B'=180^\circ$$ Therefore $A'C'B'C$ is a cyclic quadrilateral, meaning all vertices lie on a circle, which is the circumcircle of $A'B'C$. Three non-collinear points uniquely define a circle.