Question

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

$\ \pi+\sqrt{2}$

$\ \pi+2$

$\ \pi+2\sqrt{2}$

$\ 2\pi+\sqrt{2}$

$\ 2\pi+2\sqrt{2}$

Answer: B

Concepts:absolute value circle area decomposition

Difficulty rating: 1860

Solution:

The equation is symmetric in all four quadrants. In the first quadrant it becomes $x^2+y^2=x+y,$ or $(x-\tfrac12)^2+(y-\tfrac12)^2=\tfrac12.$

In the first quadrant, the enclosed region is the triangle under $x+y=1$ , with area $\frac12$ , plus a semicircle of radius $\sqrt{1/2}$ , with area $\frac\pi4$ .

Multiplying by $4$ , the total area is $4\left(\frac12+\frac\pi4\right)=2+\pi.$

Thus, the correct answer is B.

2016 AMC 10B Problem 21

Answer: B

Difficulty rating: 1860

Solution:

Problem 21 in Other Years