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.
 A: 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.
