A few months back I asked this problem. A few answers were posted;

>>8865690
>Can you do my homework?
No.

>>8865742
I don't think you grasp the nature of this problem

>>8865690
b
b for bye

>>8865770
I thought everyone was a brainlet when couldn't solve it. But kids these days don't even try. I bet you're a liberal, huh?

I've taken through calc ii and I don't have a clue how to solve this.

>>8865784
Depends what you mean by liberal

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>>8865791
one brave anon tried and got this horrible mess
(One of the probably-not-right answers)

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>a and b are constants where a < b

>>8865791
Well, as b approaches infinity, x approaches b. That's all I've got.

>>8865690
Here's the expression for the area of the intersection of the two circles

[eqn]
\begin{cases}
1 & (a>0\land x=-a-b)\lor (a>0\land x=a+b) \\
\frac{1}{2} \left(2 a^2 \tan ^{-1}\left(\frac{a^2-b^2+x^2}{2 x \sqrt{a^2-\frac{\left(a^2-b^2+x^2\right)^2}{4 x^2}}}\right)+\pi a^2+2 b^2 \tan ^{-1}\left(\frac{a^2-b^2-x^2}{\sqrt{-a^4+2 a^2 \left(b^2+x^2\right)-\left(b^2-x^2\right)^2}}\right)-\sqrt{-a^4+2 a^2 b^2+2 a^2 x^2-b^4+2 b^2 x^2-x^4}+\pi b^2\right) & a>0\land -a-b<x<a-b \\
\frac{1}{2} \left(\pi a^2-2 a^2 \tan ^{-1}\left(\frac{a^2-b^2+x^2}{\sqrt{-a^4+2 a^2 \left(b^2+x^2\right)-\left(b^2-x^2\right)^2}}\right)-2 b^2 \tan ^{-1}\left(\frac{-a^2+b^2+x^2}{\sqrt{-a^4+2 a^2 \left(b^2+x^2\right)-\left(b^2-x^2\right)^2}}\right)-\sqrt{-a^4+2 a^2 b^2+2 a^2 x^2-b^4+2 b^2 x^2-x^4}+\pi b^2\right) & a>0\land b-a<x<a+b \\
\pi (b+x)^2 & a>0\land x=a-b \\
\pi a^2 & a>0\land a-b<x<b-a
\end{cases}
[/eqn]

just set it equal to pi a^2 / 2 and solve for x

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>>8865841
>solve for x
fuck that

>>8865841
good man

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>>8865841

File: hänmänsynttävää.png (158KB, 666x607px) Image search: [Google]

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>>8865841
Fuck no.

the problem becomes trivial once the solver comes to the realization that B is actually a representation of the Earth, which is flat. Therefore, dividing A by two is as simple as aligning the flat part of B to the centroid of A, or when x = 0.

>>8865841
That very first line...
x=a+b... means there's no intersection...
Shouldn't the area of intersection be 0 then?

>>8865690
Not sure if this will work, but i'm tired, i'll try it later
I make a triangle with the circles inscribed in it, then I find the area of the missing gaps (should be 3 different ones, 5 total), then the sum of the gaps+circles-Area=area of triangle

>>8865841
be honest anon. how long did it take you to type all that up?

The solution is here:

http://mathworld.wolfram.com/Circle-CircleIntersection.html

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>>8865928
Oops, looks like just using RegionMeasure without any dimension specification caused it to think I wanted the 1-dimensional measure of the single point of intersection, which of course is 1.

Here's the actual area of the intersection, though it might be presented more simple here >>8867255
[eqn]
\begin{cases}
\pi a^2 & a-b+x<0 \\
\frac{a^4+2 \pi b^2 x \sqrt{b^2-\frac{\left(-a^2+b^2+x^2\right)^2}{4 x^2}}-2 a^2 b^2-2 a^2 x^2-4 b^2 x \sqrt{b^2-\frac{\left(-a^2+b^2+x^2\right)^2}{4 x^2}} \tan ^{-1}\left(\frac{-a^2+b^2+x^2}{\sqrt{-a^4+2 a^2 \left(b^2+x^2\right)-\left(b^2-x^2\right)^2}}\right)+\pi a^2 \sqrt{-a^4+2 a^2 b^2+2 a^2 x^2-b^4+2 b^2 x^2-x^4}-2 a^2 \sqrt{-a^4+2 a^2 b^2+2 a^2 x^2-b^4+2 b^2 x^2-x^4} \tan ^{-1}\left(\frac{a^2-b^2+x^2}{\sqrt{-a^4+2 a^2 \left(b^2+x^2\right)-\left(b^2-x^2\right)^2}}\right)+b^4-2 b^2 x^2+x^4}{2 \sqrt{-a^4+2 a^2 b^2+2 a^2 x^2-b^4+2 b^2 x^2-x^4}} & a-b+x>0\land a+b-x>0
\end{cases}
[/eqn]

>>8865928
0 seconds.
[math]\texttt{% // TeXForm // CopyToClipboard}[/math]

I swear, you faggots only know how to type shit into a calculator.

>>8867268
>1-dimensional
0-dimensional

>>8867274
A valid critique when one does not even pause to interpret the results of the calculation. But this isn't exactly a high-stakes situation...

Here's my two cents:
pia^2<pib^2
if a=1,b=2
then:
pi<4pi
half of the area of circle with radius a is pi/2
to find the area of the unknown crescent:
pi/2+A=4pi
A=7pi/2
To find the radius of x
7pi/2=pi(r)^2
7/2=r^2
r=(7/2)^(1/2)

>>8867639
now give us a general solution

>>8867783
pia^2<pib^2
if a=n, b=n+o
then
pi(n)^2<pi(n^2+2no+o^2)
half of pi(n)^2 is pi(n)^2/2
to find the area of the unknown crescent
pi(n)^2/2+A=pi(n)^2+2pi(n)(o)+o^2
A=pi(n)^2/2+2pi(n)(o)+o^2
To find radius x
pi(n)^2/2+2pi(n)(o)+o^2=pix^2
n^2+2no+o^2=x^2
x=(n^2+2no+o^2)^(1/2)
(n is some positive interger, o>0)