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Difference between revisions of "2021 April MIMC 10 Problems/Problem 18"

(Created page with "What can be a description of the set of solutions for this: <math>x^{2}+y^{2}=|2x+|2y||</math>? <math>\textbf{(A)}</math> Two overlapping circles with each area <math>2\pi</m...")
 
(→‎Solution)
 
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<math>\textbf{(E)}</math> There are two overlapping circles on the right of the <math>y</math>-axis with each area <math>4\pi</math> and the intersection area of two overlapping circles on the left of the <math>y</math>-axis with each area <math>4\pi</math>.
 
<math>\textbf{(E)}</math> There are two overlapping circles on the right of the <math>y</math>-axis with each area <math>4\pi</math> and the intersection area of two overlapping circles on the left of the <math>y</math>-axis with each area <math>4\pi</math>.
 
==Solution==
 
==Solution==
−
First, we want to graph this equation. use the technique of absolute value, there will be four cases of <math>x^{2}+y^{2}=|2x+|2y||</math>. The four cases are all circles with radius of <math>\sqrt{2}</math>. However, we realize that <math>2x</math> does not have an absolute value sign, so the left side is different from the right. Therefore, our answer would be \fbox{\textbf{(C)}}.
+
First, we want to graph this equation. use the technique of absolute value, there will be four cases of <math>x^{2}+y^{2}=|2x+|2y||</math>. The four cases are all circles with radius of <math>\sqrt{2}</math>. However, we realize that <math>2x</math> does not have an absolute value sign, so the left side is different from the right. Therefore, our answer would be <math>\fbox{\textbf{(C)}}</math>.

Latest revision as of 13:54, 26 April 2021

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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