Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$, respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$, as shown. Suppose that $AB = 2$, $O_1O_2 = 15$, $CD = 16$, and $ABO_1CDO_2$ is a convex hexagon. Find the area of this hexagon. 🔑
Solution: Let the $a$ and $b$ the radius of circle $\omega_1$ and $\omega_2$ respectively. We have $a+b=15$.
Based on Brahmagupta’s Formula, the area of cyclic quadrilateral $ABO_1O_2$ is $$S_1=\sqrt{(s-a)(s-2)(s-b)(s-15)}\tag{1}$$ where $$s=\dfrac{a+2+b+15}{2}=16$$ Simplifying (1), we have $$S_1=\sqrt{14(16+ab)}\tag{2}$$
Similarly, the area of cyclic quadrilateral $DCO_1O_2$ is $$S_2=2\sqrt{14(184+ab)}\tag{3}$$
Therefore, the area of the hexagon is $$S=S_1+S_2=\sqrt{14(16+ab)}+2\sqrt{14(184+ab)}\tag{4}$$
