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intersecting circles

There are two intersecting congruent circles, and I only know the length of the small arcs (coloured in red the image above), is there any way to find out the distance between the radii of the two circles? Thank you. :)

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    $\begingroup$ Do you know the lengths of the radii? $\endgroup$ – quasi May 21 '19 at 10:03
  • $\begingroup$ Yeah, the length is 2 units $\endgroup$ – Sophie May 21 '19 at 10:06
  • $\begingroup$ And what is the length of the red arcs? $\endgroup$ – quasi May 21 '19 at 10:12
  • $\begingroup$ the length is 2π/3 $\endgroup$ – Sophie May 21 '19 at 10:51
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enter image description here

Suppose length of the arc is l, radius of circles is r, then $r\theta = l$ where $\theta$ (in radian) is angle subtended by the arc. So $\theta = l/r$. Then the distance between center of the two circles is $2r\cos\dfrac{\theta}{2}$

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    $\begingroup$ Thank you so much! $\endgroup$ – Sophie May 21 '19 at 10:51
  • $\begingroup$ Just wondering, where is the angle subtended to? Is it the radius? I just realised the question asks for a diagram ヽ(ಠ_ಠ)ノ so what does it look like? $\endgroup$ – Sophie May 21 '19 at 11:28
  • $\begingroup$ The value you provided $l = 2\pi/3$ ; $r=2$, so $\theta = \pi/3$. $\endgroup$ – amitava May 21 '19 at 11:33
  • $\begingroup$ I'm really sorry but I'm still not sure - is 𝜃 the angle between the intersection or the one next to it? $\endgroup$ – Sophie May 21 '19 at 11:47
  • $\begingroup$ @Sophie I just added a picture in my answer, to explain. $\endgroup$ – amitava May 21 '19 at 12:03
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@amitava gave you the correct answer, for the general case of any 𝜃, r and 𝑙. In this specific case, since 𝜃 is 60° [𝜃/360°=2𝜋/3/4𝜋], we can solve it using basic geometry, without trigonometric functions.

All the radii are equal, so if you connect the centers with the endpoints of the arc, you get a rhombus. In a rhombus, the diagonals bisect the, bisect each other, and are perpendicular to each other.

So if you draw the diagonals, you get a 30-60-90 right triangle where the long leg is half of the distance between the centers, and the long leg in a 30-60-90 triangle is √3/2 times the hypotenuse. overlapping circles with rhombus

You can see a detailed step-by-step proof here:https://geometryhelp.net/distance-between-centers-overlapping-congruent-circles/

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