We have this situation:
FIGURE
We know that |CD| = 4 and |AB| = 9, CD is parallel to AB and DB touches circle in point E.
Task is to find out radius of circle.
My only idea was to somehow find right-angled triangles that will help me somehow, but without any success.
Since $CD$ and $DE$ are the two tangents to the circle from point $D$, it follows that $DE = CD = 4$. Analogously, $AB = BE = 9$. Hence $BD = BE+DE = 9 + 4 = 13$. Draw the line $DF$, where $F \in AB$, through point $D$ perpendicular to $AB$. Then triangle $BDF$ is right-angled and $ACDF$ is a rectangle, so $AC = DF$ and $AF= CD = 4$. Hence $BF = AB - AF = 9 - 4 = 5$. By Pythagoras' theorem, $$AC = DF = \sqrt{BD^2 - BF^2} = \sqrt{13^2 - 5^2} = 12$$ However, $AC$ is perpendicular to the two tangents $CD$ and $AB$, which are parallel to each other, so $AC = 12$ is the diameter of the circle. Thus, the radius is $6$.