Selection and Training Session

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 $O_1{O}_2$ and $O_{1}Z:Z{O}_2=R_1:R_2$.





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 $R_1^2-O_1{Z}^2$, 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 $ZC\cdot ZE=ZC\cdot \frac{R_1}{R_2}ZD=R_1^2-O_1{Z}^2$, i.e. $ZC\cdot ZD=\frac{R_2}{R_1}(R_1^2-O_1{Z}^2)$. Since $O_{1}Z=\frac{R_1}{R_1+R_2}{O}_1{O}_2$, 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.