Question 1151274: Suppose a circle of radius 1 is internally tangent to two circles of radius 2. What is the blue shaded area?
(please refer to the link below as it contains the image for the question)
https://i.imgur.com/l66o6SR.png
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Label the three circles A, B, C
Circle A is the circle with radius 1
Circles B and C are the circles with radius 2.
Circle A is internally tangent to circles B and C.
Place the center of circle A at the origin (0,0), which I'll call point P.
The equation of this circle is x^2+y^2 = 1
Mark two new points Q and R, such that,
Q = (-1,0)
R = (1,0)
These points are on the west and east sides of circle A, which will be the centers of circles B and C.
The equation of circle B is
(x+1)^2 + y^2 = 4
The equation of circle C
(x-1)^2 + y^2 = 4
Solve the system
to end up with the two points of intersection S and T
S = (0, sqrt(3))
T = (0, -sqrt(3))
I'll let you fill in the details for this part as an exercise. If you're stuck on how to do this, then please let me know. A hint is that you can subtract the equations and have the y^2 terms cancel out, leaving you with  .
So far we have this diagram
P = (0,0)
Q = (-1,0)
R = (1,0)
S = (0, sqrt(3))
T = (0, -sqrt(3))
Draw in triangle STR and then pull it away from the diagram. Relabel the vertices to A,B,C temporarily
the lowercase letters a,b,c are opposite their respective uppercase counterparts A,B,C.
Let's find angle C. Use the law of cosines.
c^2 = a^2 + b^2 - 2*a*b*cos(C)
(2sqrt(3))^2 = 2^2 + 2^2 - 2*2*2*cos(C)
12 = 8 - 8*cos(C)
12-8 = -8*cos(C)
4 = -8*cos(C)
cos(C) = -1/2
C = arcos(-1/2)
C = 120
This means angle SRT is 120 degrees. If we just focus on triangle SRT, then we can refer to this as "angle R" in shorthand.
Now compute the area of the blue shaded region below
which represents the circular sector TQS (centered at point R).
area of sector = (angle/360)*pi*r^2
area of sector TQS = (angle SRT/360)*pi*(QR)^2
area of sector TQS = (120/360)*pi*2^2
area of sector TQS = (1/3)*pi*4
area of sector TQS = (4/3)*pi
We will use this value later, so let M = (4/3)*pi
Compute the area of triangle SRT using the side-angle-side (SAS) triangle area rule
area of triangle = (1/2)*(side1)*(side2)*sin(included angle)
area of triangle SRT = (1/2)*(SR)*(RT)*sin(R)
area of triangle SRT = (1/2)*(2)*(2)*sin(120)
area of triangle SRT = 2*sin(120)
area of triangle SRT = 2*sqrt(3)/2
area of triangle SRT = sqrt(3)
We will use this value later, so let N = sqrt(3)
We now have this diagram
I have added the green triangle over the blue circular sector
If we take away the triangle through subtraction, then we end up with this region here
that area is exactly M-N = (4/3)*pi - sqrt(3)
as its the area of the circular sector minus the area of the triangle.
Double this result,
2*(M-N) = 2*((4/3)*pi - sqrt(3))
2*(M-N) = (8/3)*pi - 2sqrt(3)
this expression represents the exact area of this blue region shown below
The doubling can be done due to the symmetry along the y axis.
Finally, the last step is to subtract off the area of the inner circle (circle A) that has radius 1. The area of this small circle is pi*r^2 = pi*1^2 = pi square units.
So we have,
2*(M-N) - (area of circle A) = (8/3)*pi - 2sqrt(3) - pi
2*(M-N) - (area of circle A) = (8/3)*pi - 2sqrt(3) - 3pi/3
2*(M-N) - (area of circle A) = (5/3)*pi - 2sqrt(3)
This represents the final shaded blue region we are after
Therefore, the final answer is choice B
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