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Say I have a circle $x^2+y^2=R^2$ and a line $2ax+2by=R^2$ ($a,b>0$). How might I go about measuring the area of the smaller part of the circle cut off by the line?

(This question is relevant to Calculate the volume between $z=x^2+y^2$ and $z=2ax+2by$)

Thanks!

P.S. I'm not sure if this question is correctly tagged. Please correct me if I tagged it wrongly.

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    Ok, that doesn't change much for my solution below. The interpretation however is then that if the computations in my answer only have complex solutions then the line does not pass through the circle.2012-05-27

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I would start by rotating the circle such that the given line becomes parallel to the $x$-axis. That way it just boils down to integrating the function $y=\sqrt{R^2-x^2}-c$ in some range for some constant $c$. In details:

How far is the line away from the origin? The line through the origin, perpendicular to the given line is parametrised by $y=\frac{b}{a}x$. It intersects the given line at $x=\frac{aR^2}{2(b^2+a^2)}$ and $y=\frac{bR^2}{2(a^2+b^2)}$. Hence the distance is $c=\frac{R^2}{2\sqrt{a^2+b^2}}$.

Edit: If $a=0$, the above computation makes obviously no sense, but then we don't have to perform it anyway since the line is parallel to the $x$-axis in the first place.

So the area is just the integral

$\int_{-x_0}^{x_0}\sqrt{R^2-x^2}-c\ dx$

To find the suitable range of integration, we have to find the zeros of $\sqrt{R^2-x^2}-c$. They are given by $x_0=\pm\sqrt{R^2-\frac{R^4}{4(a^2+b^2)}}.$

This is all a bit sketchy, but I am sure that you can figure out the details. If not feel free to ask.

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    Oh, I see. That is indeed true. Thank you for your answer!2012-05-27
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There's a classic trick to do this with just elementary geometry:

The desired region is a wedge minus a triangle

The green region is a triangle, and the shaded region is a circular wedge: both have standard formulas for their area. To find the red region, subtract.

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    @Babak: I probably used MSPaint2012-12-08