# 2021 April MIMC 10 Problems/Problem 18

What can be a description of the set of solutions for this: $x^{2}+y^{2}=|2x+|2y||$?

$\textbf{(A)}$ Two overlapping circles with each area $2\pi$.

$\textbf{(B)}$ Four not overlapping circles with each area $4\pi$.

$\textbf{(C)}$ There are two overlapping circles on the right of the $y$-axis with each area $2\pi$ and the intersection area of two overlapping circles on the left of the $y$-axis with each area $2\pi$.

$\textbf{(D)}$ Four overlapping circles with each area $4\pi$.

$\textbf{(E)}$ There are two overlapping circles on the right of the $y$-axis with each area $4\pi$ and the intersection area of two overlapping circles on the left of the $y$-axis with each area $4\pi$.

## Solution

First, we want to graph this equation. use the technique of absolute value, there will be four cases of $x^{2}+y^{2}=|2x+|2y||$. The four cases are all circles with radius of $\sqrt{2}$. However, we realize that $2x$ does not have an absolute value sign, so the left side is different from the right. Therefore, our answer would be $\fbox{\textbf{(C)}}$.