$P_{1} O P_{2}$ and the three other analogous triangles is the area of $P_{1} P_{2} P_{3} P_{4}$. Note that if $A \neq P_{1}, P_{2}$, without loss of generality say $\angle O A P_{1}$ $90^{\circ}$, we see that $\angle O O_{2} P_{2}=2 \angle O A P_{1}$ as well, and it follows that $O O_{1} P_{1} \sim O O_{2} P_{2}$. This is a spiral similarity, so $O O_{1} O_{2} \sim O P_{1} P_{2}$, and in particular $\angle P_{1} O P_{2}=\angle O_{1} O O_{2}$, which is fixed. By the sine area formula, to maximize $O P_{1} \cdot O P_{2}$, which is bounded above by the diameters $2\left(O O_{1}\right), 2\left(O O_{2}\right)$. In a similar way, we want $P_{3}, P_{4}$ to be diametrically opposite $O_{3}, O_{4}$ in their respective circles.
When we take these $P_{i}$, we indeed have $A \in P_{1} P_{2}$ and similarly for $B, C, D$, since $\angle O A P_{1}=\angle O A P_{2}=$ $90^{\circ}$. To finish, the area of the quadrilateral is the sum of the areas of the four triangles, which is
$$
\frac{1}{2} \cdot \frac{\sqrt{2}}{2} \cdot 2^{2} \cdot(1 \cdot 2+2 \cdot \sqrt{2}+\sqrt{2} \cdot 2+2 \cdot 1)=8+4 \sqrt{2} .
$$
9. Let $O, O_{1}, O_{2}, O_{3}, O_{4}$ be points such that $O_{1}, O, O_{3}$ and $O_{2}, O, O_{4}$ are collinear in that order, $O O_{1}=$ $1, O O_{2}=2, O O_{3}=\sqrt{2}, O O_{4}=2$, and $\measuredangle O_{1} O O_{2}=45^{\circ}$. Let $\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}$ be the circles with respective centers $O_{1}, O_{2}, O_{3}, O_{4}$ that go through $O$. Let $A$ be the intersection of $\omega_{1}$ and $\omega_{2}, B$ be the intersection of $\omega_{2}$ and $\omega_{3}, C$ be the intersection of $\omega_{3}$ and $\omega_{4}$, and $D$ be the intersection of $\omega_{4}$ and $\omega_{1}$, with $A, B, C, D$ all distinct from $O$. What is the largest possible area of a convex quadrilateral $P_{1} P_{2} P_{3} P_{4}$ such that $P_{i}$ lies on $O_{i}$ and that $A, B, C, D$ all lie on its perimeter?