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I'm having difficulties finding the are of a section on the 4th circle when 4 circles intersect. The circles have a diameter of 150 mm, and the centers of adjacent circles are 100 mm apart. The shaded area is what I'm interested in. Thank you! enter image description here

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  • $\begingroup$ I guess they are adjacent 100 mm apart in form of a square? Also, why 4 circles? Aren't 3 enough? $\endgroup$ – user88595 May 22 '14 at 15:18
  • $\begingroup$ The shaded area isn't the intersection! $\endgroup$ – Ruslan May 22 '14 at 15:27
  • $\begingroup$ @user88595 That's right, the center of each adjacent circle is 100 mm apart, so we could have a 100x100 square that has 4 quarter circles. Also you're right. Only 3 out of 4 actually intersect. The reason I asked the question is because I would also like to know how to calculate the same area if the distance between the circles was 30 mm instead of 100mm. Then all 4 would intersect. I apologize for any confusion. $\endgroup$ – M731 May 22 '14 at 15:28
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Finding the intersection points between any two of the circles is not a big deal, by analytical geometry or trigonometry. Nor is finding the delimiting angles of the arcs around their respective centers.

Let two circles $(x_c, y_c, r)$ and $(x'_c, y'_c, r')$. Plug the parametric equation of the first into the implicit equation of the second:

$$(x_c+r\cos\theta-x'_c)^2+(y_c+r\sin\theta-y'_c)^2={r'}^2,$$ or after rearranging,$$2r\Delta_x\cos\theta+2r\Delta_y\sin\theta={r'}^2-r^2-\Delta_x^2-\Delta_y^2.$$

This trigonometric equation has the form $a\cos\theta+b\sin\theta=c$ and is solved by transforming it to$$\cos(\theta-\arctan\frac ba)=\frac c{\sqrt{a^2+b^2}}.$$

You will find the shaded area by using Green's theorem for areas: http://en.wikipedia.org/wiki/Green%27s_theorem#Area_Calculation.

For a single arc, the contribution to the integral is $$\frac12\oint_{Arc} x\ dy-y\ dx=\frac r2\int_{\theta_0}^{\theta_1} (x_c\cos\theta+y_c\sin\theta+r)\ d\theta=\frac r2\ (x_c\sin\theta-y_c\cos\theta+r\theta)\vert_{\theta_0}^{\theta_1}.$$

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  • $\begingroup$ If this is not detailed enough, let me know. $\endgroup$ – Yves Daoust May 22 '14 at 15:44
  • $\begingroup$ Thanks for the response. Is theta the angle that forms when I trace a line from the center of the top circle to the intersection points? I'm also a little confused about the contribution of a single arc to the area. Is this the arc in the top circle that I would then subtract from the bottom circle? I'm sorry if this sounds stupid, I'm not great at math and this is a question that my boss has brought up to me. $\endgroup$ – M731 May 22 '14 at 16:57
  • $\begingroup$ For $\theta$, it relates to the circle with center $c$ (as opposed to the one with center $c'$). It is up to you to index the circles as you wish and find the right intersections. $\endgroup$ – Yves Daoust May 22 '14 at 17:01
  • $\begingroup$ The curvilinear integral is the sum of three partial integrals for the three arc involved. For all three arcs, compute $x_c, y_c, r, \theta_0, \theta_1$ and apply the formula. $\endgroup$ – Yves Daoust May 22 '14 at 17:02

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