Is it possible that
>   Area = (1/2)(CBD)r1^2 - (1/2)r1^2*sin(CBD)
>        + (1/2)(CAD)r0^2 - (1/2)r0^2*sin(CAD)
Simplifies to what they got at
http://mathworld.wolfram.com/Circle-CircleIntersection.html at (11) ?

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> -----Original Message-----
> From: pic microcontroller discussion list
> [mailto:PICLIST@MITVMA.MIT.EDU] On Behalf Of Robert B.
> Sent: 2004 Jul 30, Fri 16:46
> To: PICLIST@MITVMA.MIT.EDU
> Subject: Re: [ee]: Compute area of overlap between two circles.
>
> From the Math Forum:
>
> I will describe a figure which you should draw out, and then
> follow my working making reference to the figure.
>
> Let A be the center of the circle (x0,y0) and B be the center
> of the other circle (x1,y1).
>
> Draw the circles with appropriate radii r0 and r1 so that
> there is a reasonable amount of overlap.  The length AB is
> calculated from the coordinates of the centers:
>
>   AB = sqrt{(x1-x0)^2 + (y1-y0)^2}
>
> For convenience let this length be denoted by c.
>
> The two circles intersect in two points which I will label C and D.
> Now we must calculate the angles CAD and CBD, and we do this
> using the cosine formula. In fact it is half of these angles
> that we first calculate, using triangle CAB.
>
>       r0^2 = r1^2 + c^2 - 2*r1*c*cos(CBA)
>            .
>            .
>   cos(CBA) = (r1^2 + c^2 - r0^2)/(2*r1*c)
>
> Having found CBA, then CBD = 2(CBA).
>
> Similarly,
>
>   cos(CAB) = (r0^2 + c^2 - r1^2)/(2*r0*c)
>
> and then   CAD = 2(CAB)
>
> Express CBD and CAD in radian measure. Then we find the
> segment of each of the circles cut off by the chord CD, by
> taking the area of the sector of the circle BCD and
> subtracting the area of triangle BCD.
> Similarly we find the area of the sector ACD and subtract the
> area of triangle ACD.
>
>   Area = (1/2)(CBD)r1^2 - (1/2)r1^2*sin(CBD)
>        + (1/2)(CAD)r0^2 - (1/2)r0^2*sin(CAD)
>
> Remember that for the area of the sectors you must have CBD
> and CAD in radians.
>
> -Doctor Anthony
>  The Math Forum
>
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