Circles $\omega_1$, $\omega_2$, and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two.  Points $P_1$, $P_2$, and $P_3$ lie on $\omega_1$, $\omega_2$, and $\omega_3$ 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 $P_4 = P_1$.  See the figure below.  The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$.  What is $a+b$?

$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$
Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$ respectively and draw $O_1O_2$, $O_1P_1$, and $O_2P_2$. Note that $\angle{O_1P_1P_2}$ and $\angle{O_2P_2P_3}$ are both right. Furthermore, since $\triangle{P_1P_2P_3}$ is equilateral, $m\angle{P_1P_2P_3} = 60^\circ$ and $m\angle{O_2P_2P_1} = 30^\circ$. Mark $M$ as the base of the altitude from $O_2$ to $P_1P_2.$ Since $\triangle P_2O_2M$ is a 30-60-90 triangle, $O_2M = 2$ and $P_2M = 2\sqrt{3}$. Also, since $O_1O_2 = 8$ and $O_1P_1 = 4$, we can find $P_1M = \sqrt{8^2 - (4 + 2)^2} = 2\sqrt{7}$. Thus, $P_1P_2 = P_1M + P_2M = 2\sqrt{3} + 2\sqrt{7}$. This makes \[\left[P_1P_2P_3\right] = \frac{\sqrt 3}4\cdot\left(2\sqrt 3 + 2\sqrt 7\right)^2 = 10\sqrt 3 + 6\sqrt 7 = \sqrt{300} + \sqrt{252}.\] So, our answer is $252 + 300 = \boxed{\textbf{D) }552}$.
Note by diyarv:
A way to see that $P_1M = 2{\sqrt 7}$ is through similar triangles. Call the intersection between $P_1M$ and $O_1O_2$ $S$. Since angles $\angle{O_1P_1S}$ and $\angle{O_2MS}$ are both right angles, and $\angle{P_1SO_1}$ and $\angle{MSO_2}$ are congruent, triangles $\triangle{P_1SO_1}$ and $\triangle{MSO_2}$ are similar by AA similarity. And, since $O_1P_1$ is $4$ and $O_2M$ is 2, the common ratio is $2$. Using the fact that $O_1S + SO_2 = 8$, we see that these lengths are $\frac{16}{3}$ and $\frac{8}{3}$ respectively. Using the Pythagorean theorem, we see that $P_1S$ and $SP_2$ are $\frac{2\sqrt{7}}{3}$ and $\frac{4\sqrt{7}}{3}$ respectively, giving us a sum of $2\sqrt{7}$.