Let $O_1$ be the center of $\Gamma_1$ and $R_1$ be its radius. Let $O_2$ be the center of $\Gamma_2$ and $R_2$ be its radius. Homothety with center at $A$ and with a positive coefficient transforms $\Gamma_1$ to $\omega_1$. Homothety with center at $B$ and with a negative coefficient transforms $\omega_1$ to $\Gamma_2$. Let $Z$ be the center of the homothety with the negative coefficient which transforms $\Gamma_1$ to $\Gamma_2$. By the theorem of three homotheties, $Z$ lies on the line $AB$. Similarly, $Z$ lies on the line $CD$. It is evident that $Z$ lies on the segment $O_1O_2$ and $O_1Z : ZO_2 = R_1 : R_2$.

Show that $ZB \cdot ZA = ZC \cdot ZD$. Let $E$ be the intersection point (different from $D$) of the line $CD$ and $\Gamma_1$. The power of point $Z$ with respect to $\Gamma_1$ is equal to $R_1^2 - O_1Z^2$, on the other hand, it is equal to $ZC \cdot ZE$. From homothety's properties we have $EZ = \frac{R_1}{R_2} ZD$. Then $ZC \cdot ZE = ZC \cdot \frac{R_1}{R_2} ZD = R_1^2 - O_1Z^2$, i.e. $ZC \cdot ZD = \frac{R_2}{R_1}(R_1^2 - O_1Z^2)$. Since $O_1Z = \frac{R_1}{R_1+R_2} O_1O_2$, we have
$$ZC \cdot ZD = \frac{R_2}{R_1} \left( R_1^2 - R_1^2 \frac{O_1 O_2^2}{(R_1 + R_2)^2} \right) = R_1 R_2 \left( 1 - \frac{O_1 O_2^2}{(R_1 + R_2)^2} \right).$$
Similarly, we obtain that $ZB \cdot ZA = R_1 R_2 \left( 1 - \frac{O_1 O_2^2}{(R_1 + R_2)^2} \right)$. Therefore, if $A$, $B$, $C$, and $D$ do not lie on the same line, then they lie on the same circle.