Solve this

[blackmage_nuke]

OK i say 6 (possibly 4) assuming you can refuel the moment all you're fuel runs out
3 planes set off at the same time

at 1/8 of the distance around the world plane 3 fuels 1 and 2 to maximum
so the fuel in each plane is now
plane 1: full
plane 2: full
plane 3: 1/4 (1/4 to get there, 1/4 to plane 1 and 1/4 to plane 2 leaves 1/4 remaining)
plane 3 has enough fuel to fly back to the airport

at 1/4 distance around the world plane 2 refuels plane 1
plane 1: full
plane 2: 1/2 (1/4 to get there from the previous refuel, 1/4 to plane 1)
thats enough for plane 2 fly back to base

when plane 1 is halfway across the world three planes (4,5 and 6) set out from the airport in the oppsoite direction
when plane 4, 5 and 6 reach 7/8 distance around the world (from the original direction) 4 refuels 5 and 6 so
4: 1/4
5: full
6: full
4 flies back to base

at 3/4 around the world (from the original direction) 5 and 6 (which now have 3/4 fuel) will meet 1 (which now has 0 fuel)
5 and 6 both give plane 1, 1/4 of their fuel each.
1: 1/2
5: 1/2
6: 1/2

This is enough fuel to get all 3 planes back to the airport

alternatively if refueling is allowed at the airport then planes 2 and 3 can be used in place of 5 and 6 meaning only 4 planes will be needed.

also:

if you fly a helper plane 1/4 the way down, it can still only give the other plane 1/4 a tank, because the other plane hasn't used more than that yet.

wouldnt a quater of the way around be half a tank since halfway around is a full tank? Not that it matters since my first answer wouldve caused the first and third plane to crash, but it's all for the greater good!

also

Thanks Tav uhmmmm how about this though:

These are two different circles. One has a fixed radius of 50, the other I have to determine based on some other criteria that isn't relevant at this point. I should have been more clear that I need to find the x-values of the points where the two circles intercept! From what I understand, I have to substitute my solved y value into the other equation, giving me...

x² + (sqrt(-x²+2500)+50)² = r²

and now solve for x... Ew.

then my first solution is what you want
x² + (y-50)² = 50² equation 1
x² + y² = r² equation 2

x² = 50²-(y-50)²
sub into equation 2
50² - (y - 50)² + y² = r²
expand the bracket
50² - (y² - 100y + 2500) + y² = r²
100y = r²
y = r²/100
sub into 2
x² + r²/100 = r²
x² = r² - r²/100
x² = 99r²/100

x = +-√(99r²/100)
x = +-(√(99)r/10)

so the circles intersect at
(√(99)r/10 , r²/100) and (-√(99)r/10 , r²/100)

The circles intercept at
x=√(99)r/10
and
x=-√(99)r/10

My bad (what trav says below)