Let $\displaystyle O_1, \ O_2$ be the centers of the circles. (The bigger circle has the center $\displaystyle O_1$), $\displaystyle A\in(O_1), \ B\in(O_2)$
Then $\displaystyle O_1A\perp AB, \ O_2B\perp AB$.
Let $\displaystyle BC\parallel O_1O_2, \ C\in (O_1A)$
In the right triangle CAB apply Pitagora:
$\displaystyle BC^2=AC^2+AB^2\Rightarrow 400=(11-r)^2+19^2$
Now solve the quadratic. Remember that $\displaystyle r<11$
Hi magentarita,
Ok, here's what I think.
In my diagram I have drawn a line from the center of the larger gear to the tangent point of the smaller gear.
Use the Pythagorean Theorem to find its length.
$\displaystyle c^2=11^2+19^2$
$\displaystyle c=\sqrt{482}$
Find angle DBA using Arctan.
$\displaystyle \arctan \frac{19}{11}\approx 59.9314$
Angle BAC is also 59.9314 since alternate interior angles are congruent (BD and CA are parallel because we have two lines perpendicular to the same line)
Now look at triangle ABC. We know $\displaystyle c=\sqrt{482}$, a = 20, and angle BAC = 59.9314.
Use the Law of Cosines.
$\displaystyle a^2=b^2+c^2-2bc \cos A$
$\displaystyle 20^2=b^2 + (\sqrt{482})^2-2b(\sqrt{482}) \cos 59.9314$
This all boils down to
$\displaystyle b^2-22b+82=0$
Apply the quadratic formula to get your 2 results. One you have to throw away because it is bigger than the larger gear radius.
Now you have your answer.