Circles $\omega_1,\omega_2,\omega_3$ each have radius $8$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1,P_2,P_3$ lie on $\omega_1,\omega_2,\omega_3$, respectively, such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$, where $P_4=P_1$. The area of $\triangle P_1P_2P_3$ can be written as $\sqrt a+\sqrt b$ for positive integers $a$ and $b$. What is $a+b$?

1. Since the area is homogeneous of degree $2$ in the common radius, doubling the radius (from $4$ to $8$) scales the area by $4$.
2. Area $=4(\sqrt{300}+\sqrt{252})=\sqrt{4800}+\sqrt{4032}$.
3. $a+b=4800+4032=\boxed{8832}$.