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Say I have any 2 circles be tangential to each other externally, that is, the distance between their centers A and B is the sum of their radii.

Is the point of tangency C included in segment AB for all cases? How is this proven?

It seems obvious if one were to draw tangential circles using a compass, but I am unable to come up with a formal explanation or proof as to why, though maybe it could be done using analytical geometry. I haven't found a answer elsewhere, neither online nor in any of my basic geometry books at hand,on whether or not this is an actual property/theorem.

SirGaethelWulf
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    With these problems it is always a problem what you can assume and what exactly you are proving. Maybe this will work for you: you can certainly find a point $C$ on $AB$ such that $AC$ is one radius (say, $r_A$). (Because $AC=r_A+r_B>r_A$.) Then $BC=AB-AC=r_A+r_B-r_A=r_B$. Thus, the point $C$ belongs to both circles. Now draw the perpendicular $l$ to $AB$ at $C$: this line is perpendicular to both radii, so it is the tangent, Q.E.D.$\blacksquare$. Having said all that, I cannot guarantee that I have not used something that, in your axiomatic system, assumes what we've just proven! –  Jan 26 '22 at 18:16
  • @StinkingBishop - How do you prove $AB = r_A + r_B$ without knowing that the point tangency is on $\overline{AB}$? – Paul Sinclair Jan 27 '22 at 18:05
  • Mind you, $C$ in the writing above is not defined to be the touching point of the common tangent. It is defined to be the point on the half-line $[AB)$ which is at the distance $r_1$ from $A$ (and therefore on the segment $[AB]$, which is longer because it is of the length $r_1+r_2>r_1$). Then I prove that $C$ is a touching point of the common tangent. –  Jan 27 '22 at 18:08
  • @StinkingBishop - You misinterpreted my remark (which apparently you read before I rewrote it to not use $C$, since I meant it as in the question, not in your proof). The problem is the fact that $|AB| = r_A +r_B$ follows from the common point of tangency lying on $\overline{AB}$, which is what you are proving. So either you have to prove $|AB| = r_A +r_B$ without knowing the common point is on that line, or your argument is circular. – Paul Sinclair Jan 27 '22 at 18:19
  • @PaulSinclair I don’t know. It seemed to me that the OP assumed $AB=r_1+r_2$ in the very first sentence. –  Jan 27 '22 at 19:45
  • @StinkingBishop - good point. I had lost sight of that give-away definition and assumed they just meant the two circles intersected in a single point. – Paul Sinclair Jan 27 '22 at 21:36

1 Answers1

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Any tangent to a circle at a point $C$ is orthogonal to the segment joining $A$ to $C$ where $A$ is the center of the circle.

If two circles of centers $A,B$ are tangent at a point $C$, the segments $AC, BC$ are orthogonal to the shared tangent line to both circles passing through $C$. Hence $A,B,C$ are aligned as desired.

mathcounterexamples.net
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  • As Stinking Bishop indicated, it's not clear what axioms OP is working from. EG Why must the tangential circles have a "shared tangent line"? – Calvin Lin Jan 26 '22 at 19:03
  • You did not use that the circles are touching externally. –  Jan 27 '22 at 18:11