I have a circle that's off-center, but I want to find out the area of the part of the circle in the positive x and y region. Not sure how to do this because of the multiple variables involved.
Area of a portion of an arbitrarily-placed circle?
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$\begingroup$
geometry
circles
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2Divide your portion of the circle into a [circular segment](http://en.wikipedia.org/wiki/Circular_segment) and a right triangle. – 2011-12-22
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0How does that work? Can it not be done with integral? – 2011-12-22
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0It can be done with integral, but that method is much easier. For example: draw a circle with center in bottom left, mark its intersections with axes A and B, let O be point (0,0) and S center of the circle; then you've got circular segment ABS and triangles ASO and BSO. – 2011-12-22
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1@sdcvvc, why two triangles? Segment AB and triangle ASB should do it. – 2011-12-22
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0Henning's answer is probably the best you can hope for. In general this is kind of a messy problem as is the case anytime you intersect circles and squares/rectangles and try to find areas -- they just don't play that nice together. The integrals you end up with are fairly nasty (involving a lot of inverse trig functions etc). – 2011-12-22
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0Is there a picture of this? I'm afraid I'm not getting it – 2011-12-22
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0@Henning: Right, I was confused; it's triangle ASB – 2011-12-22
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0The accepted answer is incorrect; see my comment. @Henning, can you post your comment with more explanation, as it is the right answer? P.S. Don't you mean triangle $AOB$ instead of $ASB$? – 2011-12-23
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0@Rahul, Yes of course triangle AOB. I had hoped somebody else would expand this into an answer because it deserves some drawings that I'm not in a position to digitize easily. – 2011-12-23