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)}}.

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2021 April MIMC 10 Problems/Problem 18

Solution