# Area of a portion of an arbitrarily-placed circle?

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.

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Divide your portion of the circle into a circular segment and a right triangle. – Henning Makholm Dec 22 '11 at 19:47
How does that work? Can it not be done with integral? – HSE Dec 22 '11 at 19:51
It 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. – sdcvvc Dec 22 '11 at 20:04
@sdcvvc, why two triangles? Segment AB and triangle ASB should do it. – Henning Makholm Dec 22 '11 at 20:08
Henning'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). – Bill Cook Dec 22 '11 at 20:09

## 1 Answer

Here's a picture for you. The area you want is the sum of the coloured areas.

Let the circle be centered at $C = (c_x, c_y)$ with radius $r$. Using the Pythagorean theorem, it's easy to find the coordinates of $A$ and $B$, and from there the areas of triangles $OAC$ and $OBC$. For the green sector $CAB$, let's call the two marked angles $\theta_1$ and $\theta_2$; these satisfy $r\sin\theta_1 = c_x$ and $r\sin\theta_2 = c_y$ respectively. The area of the sector is then $\frac12(\theta_1 + \pi/2 + \theta_2)r^2$.

@Henning: As I was making this diagram, I felt that this dissection was easier to analyse than dividing the area into a segment $AB$ and triangle $OAB$, as one can only find the area of the segment by first finding the area of the sector and then subtracting the area of triangle $CAB$.

P.S. This will still work if $c_x$ and/or $c_y$ are negative, as the areas of the corresponding triangles will simply become negative and be subtracted from the area of the sector, giving the right answer in the end. However, it won't work if the origin $O$ lies outside the circle itself. If that's a case you need to consider, it will take some more care, but a similar approach should work.

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