A few months back I asked this problem...

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

>solve for x
fuck that

good man

...

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.

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

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

be honest user. how long did it take you to type all that up?

The solution is here:

mathworld.wolfram.com/Circle-CircleIntersection.html

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 [eqn]
\begin{cases}
\pi a^2 & a-b+x0\land a+b-x>0
\end{cases}
[/eqn]

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

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

>1-dimensional
0-dimensional

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

now give us a general solution

pia^2