smc

The circles match up as follows: Case $1$ is brown, Case $2$ is blue, Case $3$ is green, and Case 4 is red. Let $x^2+y^2=64$ be circle $O$, $x^2+y^2=4$ be circle $P$, and $(x-5{)}^2+y^2=3$ be circle $Q$. All the circles in S are internally tangent to circle $O$. There are four cases with two circles belonging to each: $\ast$ $P$ and $Q$ are internally tangent to $S$. $\ast$ $P$ and $Q$ are externally tangent to $S$. $\ast$ $P$ is externally and Circle $Q$ is internally tangent to $S$. $\ast$ $P$ is internally and Circle $Q$ is externally tangent to $S$. Consider Cases $1$ and $4$ together. Since circles $O$ and $P$ have the same center, the line connecting the center of $S$ and the center of $O$ will pass through the tangency point of both $S$ and $O$ and the tangency point of $S$ and $P$. This line will be the diameter of $S$ and have length $r_P+r_O=10$. Therefore the radius of $S$ in these cases is $5$. Consider Cases $2$ and $3$ together. Similarly to Cases $1$ and $4$, the line connecting the center of $S$ to the center of $O$ will pass through the tangency points. This time, however, the diameter of $S$ will have length $r_P-r_O=6$. Therefore, the radius of $S$ in these cases is $3$. The set of circles $S$ consists of $8$ circles - $4$ of which have radius $5$ and $4$ of which have radius $3$. The total area of all circles in $S$ is $4(5^2\pi +3^2\pi )=136\pi \Rightarrow \boxed{136\pi }$.