Tough DS

x^2+y^2=4 --> represents CIRLCE at center 0,0 and radius 2

and y=|x| --> for every value of "x", whether -ve or +ve, "y" is positive

theCodeToGMAT wrote:x^2+y^2=4 --> represents CIRLCE at center 0,0 and radius 2

and y=|x| --> for every value of "x", whether -ve or +ve, "y" is positive

I did the same as you did but then I got confused with this thing.

If you take x=4,-4 then Y=4 so the area will widen right? you have only done with x=1,-1 and y=1.
My question is why didn't you consider x=4,-4 or 5,-5 or so on. If you chose those the area will be bigger. That made me confused.

Rakesh, we need to find the smaller sector area.. Sector area means area covered inside a circle for some particular angle..

The radius of circle = 2

Now, if we consider x = -4 or 4 then correspondingly y = 4.. the corresponding overlap region is only up to x=2 & y=2 ..

So, the restricted region is defined by the circle's radius = 2

By the way, I din't consider x = 1 or -1

Graph of y = |x| is similar to the graph of y = x but with one exception : the -ive side is reflected over the x axis.
Thus for y = |x|
y = x for x > 0
y = -x for x < 0
Essentially the negative half of the y = x would be reflected up creating a "V" shape.

Also, the general equation of a circle with center at (a, b) and radius r --> (x - a)² + (y - b)² = r²
In our case a and b are zero thus the center is at the origin : x² + y² = 2²


We need the area of the common region between the V shape and the circle with a radius of 2 units (the shaded sector)

Area of the sector =

Answer C