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 which 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.)

I want a clear explanation with the answer. Thanks!

Each small circle has area pi

The intersections each have area pi/2 - 1, so the 4 areas of intersection have area 2pi-4

The large circle has area 4pi.

The points outside all the small circles lie only in the big circle.

the points in each of the intersections lie in 2 small circles and the big circle.

So, now you know which areas to add up.

To find the area of all points contained in an odd number of these circles, we can use the principle of inclusion-exclusion.

We can start by finding the total area of the five circles:
- Circle 1: Center (1,0) and radius 1
- Circle 2: Center (-1,0) and radius 1
- Circle 3: Center (0,1) and radius 1
- Circle 4: Center (0,-1) and radius 1
- Circle 5: Center (0,0) and radius 2

The area of each circle is given by A = πr^2, where r is the radius.

Now, let's consider the overlapping areas between these circles.

Circles 1 and 2 intersect at the x-coordinate -1/2, forming a straight line segment with endpoints (-1/2,-√3/2) and (-1/2,√3/2). The area of this intersection is a triangle with base of length 1 and height √3/2, giving an area of (1/2)(1)(√3/2) = √3/4 π.

Similarly, circles 3 and 4 intersect at the y-coordinate -1/2, forming another triangle with the same area of √3/4 π.

Circles 1, 3, and 5 intersect at the point (1/2, √3/2), forming another triangle with a base of length 1 and height √3/2, giving an area of √3/4 π.

Circle 2, 4, and 5 also intersect at (-1/2, √3/2) and form a triangle with the same area of √3/4 π.

Finally, all five circles intersect at the origin (0,0), forming a pentagon. To find the area of this pentagon, we can subtract the areas of the four triangles from the area of the square formed by the radius of the large circle.

The area of the square is (2r)^2 = 4^2 = 16 square units.

The combined area of the four triangles is 4 * (√3/4 π) = √3 π.

So, the area of the pentagon is 16 - √3 π.

Now, we can calculate the area of the region contained in an odd number of these circles by using the principle of inclusion-exclusion.

The area contained in an odd number of circles is equal to the sum of the areas of the individual circles minus the sum of the areas of the pairwise intersections plus the area of the pentagon formed by the intersection of all five circles:

A = Circle 1 + Circle 2 + Circle 3 + Circle 4 + Circle 5 - (Circles 1 & 2) - (Circles 3 & 4) - (Circles 1, 3 & 5) - (Circles 2, 4 & 5) + Pentagon

A = π(1)^2 + π(1)^2 + π(1)^2 + π(1)^2 + π(2)^2 - 2 * (√3/4 π) - 2 * (√3/4 π) - 2 * (√3/4 π) - 2 * (√3/4 π) + (16 - √3 π)

Simplifying this expression, we get:

A = 4π + 4π + 4π + 4π + 4π - 8√3π + 16 - √3π

A = 20π - 9√3π + 16

So, the area of all points contained in an odd number of these 5 circles is 20π - 9√3π + 16.