Applying Pythagoras' theorem to the diagram on the right gives $$\begin{align}(2r)^2+6^2=&(r+6)^2\\
\Rightarrow4r^2+36=&r^2+12r+36\\
\Rightarrow3r^2=&12r\\
\Rightarrow3r^2-12r=&0\\
\Rightarrow3r(r-4)=&0\\
\Rightarrow r-4=&0\hspace{12mm}(r\ne0)\\
\Rightarrow r=&4\end{align}$$