Given
( x + a ) ² + y ² = r ² , ( intersects y axis at ±4 )
( x₁ - a₁ ) ² + y₁² = r₁²
( x₂ + a₂ ) ² + y₂² = r₂²
r₁ + r₂ = r
The y axis is tangent to A₁ and A₂ .
Find the enclosed area A - A₁ - A₂
Edit : All of A₁ is in A .
All of A₂ is in A .
Dr D
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First things first. For the first circle, A
r^2 = a^2 + 16
For the next two circles,
r1 = a1
r2 = a2
There is no intersection between circles A1 and A2 since they are both tangential to the y axis and in different sides of it.
Circle A intersects the x axis at -a +/- sqrt(16+a^2)
The positive intersection 0 < -a + sqrt(16+a^2) < 4
Circle A1 intersects the x axis at 2*r1
If r1 > r2, then circle A2 is completely inside circle A, and part of circle A1 is inside A.
We can find the points of intersection between A and A2 by setting their equations equal to each other. They will intersect at
x = [ r*r2 + r1*r2 + r1^2 - a^2 ] / [2 *(a + r1) ]
To find the area of circle A1 that is outside of A, we can integrate 2*(y1 - y)*dx from
x = [ r*r2 + r1*r2 + r1^2 - a^2 ] / [2 *(a + r1) ]
to x = -a + sqrt(16+a^2)
Then integrate 2*y1*dx from
x = -a + sqrt(16+a^2)
to x = 2*r1
There will be a bunch of cumbersome algebra. Good luck with it.
**EDIT**
If all of A1 and A2 are inside A, then that means
a = a1 - a2
because the smaller circles will be tangential to the large one.
Other equations:
2a1 = r - a = sqrt(a^2 + 16) - a
2a2 = r + a = sqrt(a^2 + 16) + a
So r^2 - a1^2 - a2^2
= a^2 + 16 - 1/4 * (4a^2 + 32)
= 8
A - A1 - A2 = 8π