I admire you maths types but you make my head hurt. I have read this thread twice and still can’t make top nor tail of it.
Sorry to interrupt, I just wanted to say “cor you lot are clever” 
If you follow the equation exactly like it is in the link, when d=41.1009 the area=424. When d=28 then a=854.052. Can I see the equation?
It should be, using my calculator to reduce:
-729arcsin((x^2+368)/54x)-361arcsin((x^2-368)/38x)-.5sqrt(-x^3+8x^2+2116x-16928)+54pi
You raise an interesting point with radians v. degrees but I am absolutely sure that my answer of 41.1009 will yield 424. Maybe there are others but I really don’t think there would be. Then again, my answer is not 100% correct since I used 27 and 19 instead of the exact values. I will, however, use a proof using the law of consine and areas of sectors if I need to.
I tried the equation first using the exact (unsimplified with the exception of inputting R and r values) formula given by Mathworld, then by simplifying and using first 1/cos(stuff) and then by using tan(stuff)/sin(stuff). Using arccos I got exactly one integer value (a value for x=0) and using tan/sin I got plenty but no zero value for Y for 28 or 41. Using 1/cos I got the same values as for tan/sin.
Two points:
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Yes, you should be working in radians. Degrees are very rarely used in pure math.
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ZebraShaSha, I have no idea where you got that big long equation. Can you explain?
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When I graphed the equation from mathworld with plugged in r and R values my calculator was set to radians.
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I have no idea, either. Using my Ti-89 I did factor(originial equation with plugged in r values). It is not the one I used to get my answer but it is the most simplified you can get.
Shhh. They’re having fun. (No, you didn’t miss anything. They’re not doing it because they need to.)
I just checked my results with my CAD program at work. Drawing circle C1 with radius R1=19.14923 and circle C2 with radius R2=26.6904, and locating the two circles such that distance between the center points of the two circles is d=28.48349, the area of the intersection of the two circles is 424.
If the two circles are located such that d=41.1, the intersected area is about 64.
I no longer have the math skills to simplify and solve the equation, but I believe my spreadsheet method has determined at least one correct solution (d=28.48439).
Plug your answer into the equation. What do you get? I got 841.122, but maybe it is just me. Now, plug 41.1009 into the equation. What do you get?
Revising my answer, so that they are not rounded off integers, I am now getting 40.9028=d.
Using the formula in the MathWorld site, if r=19.14923 and R=26.6904:
d=28.48349, A=423.9966
d=40.9028, A=67.80075
I think the equation you are using is not equivalent to the one provided in the MathWorld site.
With d = 28.5, you get a very close area (423 and change). That’s probably good enough for government work.
:bangs head against the wall:
I was using the one off of mathworld but for some reason I forgot a major part of it. Under the square root I missed an entire binomial. After checking many, many times I seemed to lose this part. Some how, no matter what, I am destined to make major mistakes anytime I do mathematics. I also seemed to type it wrong more than once, since I typed it into my Ti-83 and twice on the Ti-89. Sorry about that.
Yes, the answer is 28.4834.
That’s OK, I make silly math mistakes all the time. And that was quite a long equation to enter.
Here’s something interesting.
I decided the easiest way to find a suitable value of d was by iteration. Take a value of d and calculate the area. If it’s too big, make d larger. If it’s too small, make d smaller.
I did that by picking an upper bound of R + r, and a lower bound of R. If the area was bigger than the target of 424, I set d to be (d + R + r)/2, and if it was smaller, (d + R)/2. The area should be a monotonic function of d, so I thought this should arrive at an answer fairly quickly.
No, it orbits four different values of d, and it settles pretty quickly. It’s an interesting system, but not so helpful for getting an answer.
Hey! Well, okay.
I’m curious…why use circles for your graph, when overlapping squares will give you the same result, and will be much easier to draw?
Puh-leeezze! :rolleyes:
I’m joking, of course, because that is such an obvious idea that I’m embarrassed to have not thought of it myself. I wonder how it’ll look… I’ll have to give it a whirl.
How’s this look?