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I'm trying to formally show that if two circles $\Gamma$ and $\Gamma'$ are tangent at $A$, then $\Gamma'$ lies either entirely inside or entirely outside $\Gamma$.

The one particular case I'm struggling with is when $\Gamma$ and $\Gamma'$ lie on the same side of the tangent line $l$ at $A$. I know that $O$, $O'$ and $A$ are collinear. I suppose the radius $r$ of $\Gamma'$ is smaller than radius $s$ of $\Gamma$, and I want to show that $\Gamma'$ is entirely inside $\Gamma$, except for $A$.

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I know line $OO'$ meets $\Gamma'$ on the other side of $A$ at some point $C$. I've been able to show that $C$ is inside $\Gamma$. For if $C*O*O'$, then $CO<CO'=r<s$, so $C$ is inside $\Gamma$. If $O*C*O'$, then $CO<OO'<OA=s$, so again $C$ is inside $\Gamma$. But when I take an arbitrary point $B$ on $\Gamma'$, I have no idea how to show that $OB<s$, to prove that $B$ is inside $\Gamma$. Does anyone know a nice way to rigorously show this? Thank you.

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Suppose a ray emanating from $A$ meets circle $O$ (of radius $r$) at point $P$ and circle $O^\prime$ (of radius $r^\prime$) at $P^\prime$. Triangles $OAP$ and $O^{\prime}AP^\prime$ are similar isosceles triangles. (Their base angles are congruent to the angle they share at $A$.) Therefore, $|AP^\prime|/|AP| = r^\prime/r$, which implies that the closer (to $A$) of $P$ and $P^\prime$ always corresponds to the smaller of $r$ and $r^\prime$, regardless of where the ray is pointing. –  Blue Feb 28 '11 at 4:08
    
@Day Late Don, thanks for your comment, it's a nice way to look at it. I'm afraid the axiomatization I'm using hasn't developed any theory of similar triangles yet, but much appreciated. –  yunone Feb 28 '11 at 4:28

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I'm not sure what axiomatization of Euclidean Geometry you're using, and what results you are considering as given at this point, but are there any reasons why you cannot use the triangle inequality to show that $OB<OO^{'}+O^{'}B$? Then since $O^{'}B=O^{'}A$, you would have that $OB<OA$. (You have claimed that I can take as given that $O$, $O^{'}$, and $A$ are colinear).

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Thank you Albert, this works very well. I'm working in a Hilbert plane, and I have shown for myself that the triangle inequality of I.20 of Euclid does hold with the Hilbert axioms I'm using. –  yunone Feb 28 '11 at 4:26

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