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In general any two circles have two centers of homothety. They have only one center when the circles have the same radius or when the circles have the same center.

Given two circles of different radii with disjoint interiors, their external center of homothety is at the intersection of the external tangents and the internal center is at the intersection of the internal tangents. This is the case because homothety preserves tangency. So the centers of homothety are easy to construct in this case.

If the circles partially overlap, there is still an external homothetyic center at the intersection of external tangents, and an internal homothety center lying somewhere along the line joining the centers of the circles. If the circles overlap completlety (one inside the other), there are two internal homothety centers on the line joining the centers of the circles.

My question is how do you construct (with compass and straightedge) the centers of homothety in the cases where the circle interiors are not disjoint.

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In the case of two circles that are not disjoint, I think that you only have one homothety center, which is the one found by the intersecction of the external tangents. The reason for that is that when you have two circles that are not disjoint the internal tangent no longer exists. In case you don't know how to find the external tangent of two cercles (it doesn't matter if they are disjoint or not) with compass and straightedge, here you have images and an explanation that shows you how to do it.

http://jwilson.coe.uga.edu/emt669/Student.Folders/Kertscher.Jeff/Essay.3/Tangents.html

Basically, you draw a line between the two centers, you find the medium point of this line and draw a circle centered there that touches both cirle centers using the compass. After that, you draw a circle centered in the center of the biggest circle with a radious equal to R= R(big circle)-R(small circle). You draw a line from the center of the biggest circle to the point that you find on the intersection of the two new circles that you've drawn, and follow this line until it intersects with the biggest circle. This point is the first tangent point. If you now draw a parallel line to this one that passes through the center of the small circle, you'll have the second tangent point in the intersection of this line with the small circle. If you draw a line joining both tangent points and make it intersect with the line that joins both circle centers, there you'll have your homothetic center.

For more information, you may want to take a look at this:

https://commons.wikimedia.org/wiki/Category:Homothety_in_circles

Maybe you are looking not for homothety centers but antihomothety centers. I hope I've been able to help you, and don't hesitate to tell me if not, i'll try to do my best!

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  • $\begingroup$ Yes I know how to construct tangent lines to circles. And I am pretty sure there are always 2 centers of homothety except in the cases I mentioned when the centers coincide or the radii are equal. Internal tangents don't have to exist for there to be a center of homothety.... However if a common tangent line does exist, one of the centers must lie on that line. $\endgroup$ – Joshua Benabou Jul 18 '15 at 20:05
  • $\begingroup$ Why do you give me a downvote if you aren't even sure of what you say? As I mention in my answer, this is what I think that happens. If you aren't satisfied with my answer, just tell me, but downvoting me isn't the way to do it I think, not if I do the effort of writting an elavorated answer... Thank you. @JoshuaBenabou $\endgroup$ – Mr.Polymath Jul 22 '15 at 11:51
  • $\begingroup$ Ummm I didn't downvote your answer.... But also you didn't do anything except repeat what I said in the OP. $\endgroup$ – Joshua Benabou Jul 22 '15 at 17:48
  • $\begingroup$ Not true since you asked how to draw it with compass and straightedge and you didn't say if you already knew how to do it or not. $\endgroup$ – Mr.Polymath Jul 23 '15 at 6:49
  • $\begingroup$ Yes I did. I said "so the centers are easy to construct in this case". $\endgroup$ – Joshua Benabou Jul 23 '15 at 6:52

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