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To-day I want to look at CCC - one circle tangent to three circles whose radii and positions of their centers are known.

How does one solve this.. old fashioned ways like ruler and compass, or modern ways like intersection theory?

these Apollonius problems are all very classical but good treatments are hard to find. here is a different such problem

construct circle tangent to two given circles and a straight line

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Here is an outline of the a procedure. Assume first that you want to find a circle that encloses all three others and touches them from the outside.

  1. Observe that if you increase the radii of all the three circles by the same amount, then the center of the surrounding circle will stay at the same point. Only its radius will increase (by the same amount). Therefore without loss of generality we can assume that two of the three circles intersect at a point $A$. After all we can achieve this by adjusting their radii.
  2. Invert the whole picture w.r.t. a circle with center $A$ (pick one, doesn't matter which one you use). Remember that inversion preserves tangency, and takes circles to circles (or to lines in an important special case!). Because two circles passed via $A$, those two circles become lines in the inversion. Now we have the task of finding a circle that is tangent to a given circle and to given two lines.
  3. Reusing the idea of the first stage we then shrink the remaining circle to have radius zero, i.e. shrink it to a point. At the same time we shrink the radii of the two lines by parallel transporting them by the same amount. Again, in this process the center of the sought after circle does not move.
  4. We now face the problem of finding a circle that is tangent to two given lines and passes through a given point. Solving this is easy. Draw any circle tangent to the two lines (its center will be on the bisector of the angle formed by the two lines et cetera). That circle probably won't go through the point, but a simple application of homothety will fix that.
  5. Undo steps 3,2 and 1 in that order. Inversion is its own inverse, and the other steps have the obvious inverse operations. Rejoice!

If you want to vary the way of tangency (inside/outside), then you may need to shrink some of the three circles instead of expanding their radii in step 1. Altogether there may be as many as eight solutions, for the fourth circle may be tangent to the other three in two ways. Some of those alternatives may not work. For example, if one of the three original circles is inside another, we get obvious problems. Also, should the three initial circles be concentric, there cannot be a solution, and in such a case already step 1 fails.

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  • $\begingroup$ reminds me of en.m.wikipedia.org/wiki/Smallest-circle_problem $\endgroup$ – cactus314 Jul 16 '16 at 22:10
  • $\begingroup$ Looks simple enough, but I shudder to think how complicated the general formulas would be (i.e. the center and radius of the tangent cirlce expressed using the centers and radii of the three initial circles) $\endgroup$ – Yuriy S Jul 16 '16 at 22:43
  • $\begingroup$ Formulas!? Not this way :-) More like a straightedge+compass construction. $\endgroup$ – Jyrki Lahtonen Jul 17 '16 at 8:09
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THIS IS A COMMENT NOT AN ANSWER.

In http://debart.pagesperso-orange.fr/seconde/contruc_cercle.html there are ten problems of contact concerning circles which are solved in a quite interesting way. The text is in French and the problem $10$ is the Apollonius’s consisting in finding a circle tangent to three given circles of which eight solutions are given.

Here a translation of the “Historique” of the subject. $$*****$$

“ The problem of the circle tangent to three circles is one of the great problems of history of geometry.

It was introduced by Pappus as the tenth and most difficult of the Treaty of contacts, one of the lost works of Apollonius.

In 1596, Adrien Romain (Van Roomen, Latinized as Adrianus Romanus, Flemish mathematician 1561-1615) propose a solution using a hyperbola, that Vieta considers not in accordance with the method of Ancients.

Indeed, the emergence of algebra in geometry allows him to claim that it is a problem of the second degree, so a plan problem which can be solve with "the ruler and compass."

Vieta published his own solution in 1600, in his Apollonius Gallus where he presents some lemmas for handling the similarities and where he exhibited the first nine situations listed above. It recognizes that the solutions to this problem tenth depend on the relative position of the three circles, but ignores the discussion of a lot of solutions and will have to wait until Descartes (1637) for discussion of particular solutions.

Until the nineteenth century, this problem will be one of the places of confrontation between the synthetic geometry (pure geometry) and analytical geometry."

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  • $\begingroup$ Thank you for this useful historical remark. Even though it's not an answer, you get my +1 $\endgroup$ – Yuriy S Jul 16 '16 at 22:44
  • $\begingroup$ Very kind, thank you. The "Historique" I translated, precedes the eight solutions of the tenth and last problem in the mentioned source. And I did it because many do not know French and the prologue to the problem I found very interesting. $\endgroup$ – Piquito Jul 16 '16 at 23:41

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