2022 AMC 10B Problem 22

Call the concentric circles of radii $2$ and $8$ the inner and outer circles. Let a desired circle have radius $r$ and let its center be distance $d$ from the origin. It must be internally tangent to the outer circle, so $d+r=8.$ If it is externally tangent to the inner circle, then $d-r=2,$ giving $(r,d)=(3,5).$ If it contains the inner circle, then $r-d=2,$ giving $(r,d)=(5,3).$ Thus every desired circle has radius $3$ or $5.$ The third given circle has radius $\sqrt3$ and center $(5,0).$ For either value of $r,$ a desired circle may be internally or externally tangent to it, so the distance between their centers is $r-\sqrt3$ or $r+\sqrt3.$ In all four cases, if this distance is $q,$ then $|d-q|<5<d+q,$ so the circle of possible centers intersects the circle of radius $d$ about the origin in two points, symmetric across the $x$-axis, as shown. Hence there are $4$ desired circles of each radius. The total area is therefore $$4(5^2\pi+3^2\pi)=136\pi.$$ Thus, E is the correct answer.