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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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Do you have any conditions on $a$ and $b$ (relative to each other or to $R$)? –  Simon Markett May 27 '12 at 12:20
    
@SimonMarkett: Sorry for taking a while to notice your comment. I forgot to mention that $a, b > 0$, thanks. –  ro44 May 27 '12 at 14:45
    
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. –  Simon Markett May 27 '12 at 14:56
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2 Answers

up vote 1 down vote accepted

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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The one part I am unclear about is, how would I go about performing the rotation? Thanks! –  ro44 May 27 '12 at 14:57
    
I am not literally performing a rotation, that is not algebraically. The observation is that the we get the same result if we take a different line, namely the one parallel to the $x$-axis, which has the same distance to the origin as the given one. This is true since this is only a rotation geometrically. –  Simon Markett May 27 '12 at 14:59
    
Oh, I see. That is indeed true. Thank you for your answer! –  ro44 May 27 '12 at 15:01
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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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Nice trick! Thank you. –  ro44 May 27 '12 at 15:19
    
I thought about doing it that way. But with the given data computing the angle and the measurements of the triangle didn't seem that easy after all. (In theory it is easy, but in practice the formulas don't seem to simplify much.) –  Simon Markett May 27 '12 at 15:38
    
@Simon: No, it is easy! You compute the distance of the line from the center of the circle, and then the angles follow from simple trigonometry. –  Rahul May 27 '12 at 17:28
    
Nice! Nice!. how could you colored that bounded areas? Did you use any certain software? –  B. S. Dec 8 '12 at 10:05
    
@Babak: I probably used MSPaint –  Hurkyl Dec 8 '12 at 10:38
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