Area of region | AMC 10B, 2016| Problem No 21

Pre College Mathematics

AMC-10B, 2016 Problem-21

$\pi+2$

The given equation is $x^{2}+y^{2}=|x|+|y|$. Expanding this equation we get four equation as mod exist here...

$x^2+y^2-x-y=0$.......................(1)

$x^2+y^2+x+y=0$....................(2)

$x^2+y^2-x+y=0$.....................(3)

$x^2+y^2+x-y=0$.....................(4)

using this four equation can you draw the figure ?

Now can you finish the problem?

now four equations can be written as $x^{2}-x+y^{2}-y=0 \Rightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$

$x^{2}+x+y^{2}+y=0 \Rightarrow\left(x+\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$

$x^{2}-x+y^{2}+y=0 \Rightarrow\left(x-\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$

$x^{2}+x+y^{2}-y=0 \Rightarrow\left(x+\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}=\left(\frac{\sqrt{2}}{2}\right)^{2}$ which represents four circles and they overlapping.....

The center of the four circles are $\left(\frac{1}{2}, \frac{1}{2}\right)$, $\left(\frac{-1}{2}, \frac{-1}{2}\right)$,$\left(\frac{1}{2}, \frac{-1}{2}\right)$,$\left(\frac{-1}{2}, \frac{1}{2}\right)$Now we have to find out the region union of the four circles.

Now can you finish the problem?

There are several ways to find the area, but note that if you connect (0,1),(1,0),(-1,0),(0,-1) to its other three respective points in the other three quadrants, you get a square of area 2 , along with four half-circles of diameter $\sqrt{2}$, for a total area of $2+2 \cdot\left(\frac{\sqrt{2}}{2}\right)^{2} \pi=\pi+2$