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I have a circle with a known center and radius. I have an arc with a known radius and two points on its edge, where one end is at the same position as the circle's center. Is there a way to find their point of intersection that's simpler than turning this into two circles and intersecting those (and somehow figuring out which of the two points of intersection to use)?

In other words, the green bits of this diagram are known, and I'm trying to find the red:

Circle and Arc Intersection

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  • $\begingroup$ Your drawing suggests that the centres of the circle and the arc have the same y co-ordinate. Did you intend this ? $\endgroup$
    – WW1
    Jul 31, 2016 at 0:02
  • $\begingroup$ Ah, no I did not. Apologies for the ambiguity! The center of the arc can be anywhere, though actually in my case, the arc will always curve counter-clockwise of the circle (though that probably makes no difference). $\endgroup$
    – Yousef Amar
    Jul 31, 2016 at 0:18
  • $\begingroup$ There are two points of intersection, depending on where you put the center of your arc. Even considering that the arc's curve is "counter-clockwise," you still can place the arc's center on either side; it just flips if the arc is coming out of or going into the circle. $\endgroup$
    – Carser
    Jul 31, 2016 at 3:45

1 Answer 1

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enter image description here

From the given, I assume that J(h, k) can be found. Then, equation of OJ will be known. (This is especially simple if the y-ordinates of O and J are the same according to your drawing).

Applying the cosine law to the R-r-R triangle, $\theta$ can be found.

OX is the line that deviates $\theta$ from the line OJ. X is on the intersection of that line and the red circle.

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    $\begingroup$ I often say most problems in geometry can be easily solved once the correct triangle is pointed out. This is one of those cases. $\endgroup$
    – John Alexiou
    Jul 31, 2016 at 5:48
  • $\begingroup$ @ja72 Thankyou. $\endgroup$
    – Mick
    Jul 31, 2016 at 11:23

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