Four circles of unit radius are drawn with centers $(1,0)$, $(-1,0)$, $(0,1)$, and $(0,-1)$. a circle with radius 2 is drawn with the origin as its center. what is the area of all points that are contained in an odd number of these 5 circles? (express your answer in the form "a pi + b" or "a pi - b", where a and b are integers.)

Respuesta :

Check the picture 1.

The regions contained in an odd number of these 5 circles are

the pink colored region, contained only in the circle with radius 2, 

and the yellow regions, each contained in the intersection of 2 small circles and the third large circle.

So what we need to determine is the "area pink + area yellow".


Consider the second picture.

Let A2 be the area of the right isosceles triangle, and A1 be half of the yellow region.

the region A2 + A1 is called a sector, and it is 1/4 of the area of a unit circle.

[tex]A1=A_{sector}-A2= \frac{1}{4} \pi r^2- \frac{1}{2}.1.1= \frac{1}{4} \pi 1^2- \frac{1}{2}= \frac{1}{4}\pi- \frac{1}{2}[/tex]


Thus, the overall yellow region is [tex]8A1=8(\frac{1}{4}\pi- \frac{1}{2})=2\pi-4[/tex]


The purple area is:
 
Area of large circle - 4*(Area of 1 small circle) + yellow region.

(because subtracting all 4 small circles means that each of the 4 separate yellow regions have been subtracted twice)

[tex]A_{purple}=\pi.2^2-4(\pi.1^2)+2\pi-4=4\pi-4\pi+2\pi-4=2\pi-4[/tex]


Finally, the total area is : [tex]2\pi-4+2\pi-4=4\pi-8[/tex] (units squared) 



Answer:   [tex]6\pi-8[/tex]    square units
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