Why are tangent circles entirely inside or outside the other?

Question

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 $A$.

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.

Answer 1

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).