smc

We are given that $BC$ is tangent to the smaller circle. Using that, we know where the circle intersects $BC$, it creates a right triangle. We can also point out that since $AC$ is the diameter of the bigger circle and triangle $ABC$ is inscribed the semi-circle, that angle $B$ is a right angle. Therefore, we have $2$ similar triangles. Let's label the center of the smaller circle (which is also the center of the larger circle) as $D$. Let's also label the point where the smaller circle intersects $BC$ as $E$. So $ABC$ is similar to $DEC$. Since $DE$ is the radius of the smaller circle, call the length $x$ and since $DC$ is the radius of the bigger circle, call that length $3x$. The diameter, $AC$ is $6x$. So, $\frac{AB}{AC}=\frac{DE}{DC}\Rightarrow \frac{12}{6x}=\frac{x}{3x}\Rightarrow 36x=6x^2\Rightarrow 6=x$ But they are asking for the larger circle radius, so $3x=\boxed{18}$ $\boxed{18}$