2021 AMC 10A Problem 19

Consider the four sign cases for $x-y$ and $x+y.$ In one case, for example, $|x-y|=x-y$ and $|x+y|=x+y,$ so $$\begin{gathered} x^2+y^2=6x \\ \quad\Longrightarrow\quad (x-3)^2+y^2=9. \end{gathered}$$ The other three cases similarly give circles of radius $3$ centered at $(0,3),$ $(-3,0),$ and $(0,-3).$ The relevant arcs form the boundary shown by these four congruent circle pieces. The region consists of a central square of side length $6,$ together with four semicircles of radius $3.$ The square contributes area $36,$ and the four semicircles have the area of two full radius-$3$ circles, namely $18\pi.$ Therefore the area is $36+18\pi,$ so $m+n=36+18=54.$ Thus, E is the correct answer.