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AMC10 2016b Test Paper
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Sample Question 21:

What is the area of the region enclosed by the graph of the equation \(x^2+y^2=|x|+|y|?\)

\(\textbf{(A)}\ \pi+\sqrt{2}\qquad\textbf{(B)}\ \pi+2\qquad\textbf{(C)}\ \pi+2\sqrt{2}\qquad\textbf{(D)}\ 2\pi+\sqrt{2}\qquad\textbf{(E)}\ 2\pi+2\sqrt{2}\)




Answer Keys

Question 21: B


Solutions

Question 21
Step 1: Begin by considering only the first quadrant. Since the equation \(x^2+y^2=|x|+|y|\) is symmetric for all four quadrants, we can later multiply by 4 to get the total area for all quadrants.

Step 2: We work within the constraints of the first quadrant where \(x, y >= 0\). This simplifies the equation as \(|x|=x\) and \(|y|=y\).

Step 3: Rewriting the equation we get \(x^2-x+y^2-y=0 \rightarrow (x-\frac{1}{2})^2+(y-\frac{1}{2})^2=(\frac{\sqrt2}{2})^2\). This is the equation of a circle with center \((\frac{1}{2}, \frac{1}{2})\) and radius \(\frac{\sqrt2}{2}\).

Step 4: The area enclosed by this circle in the first quadrant can be obtained by adding the area of a square of side 1 and the area of half a circle with radius \( \frac{\sqrt2}{2} \). Therefore, the area is \( 1 + \frac{1}{2} \cdot \pi \cdot (\frac{\sqrt2}{2})^2 = 1 + \frac{1}{4}\pi \).

Step 5: Since the equation is symmetric for all four quadrants, multiply this area by 4. This results in \( 4 + \pi \), which matches answer choice B: \(\pi + 2\).