smc

First, observe that the two tangent lines are of identical length. Therefore, supposing that the point of intersection is $(x,0)$, the Pythagorean Theorem gives $\sqrt{(x-6{)}^2+13^2}=\sqrt{(x-12{)}^2+11^2}$. This simplifies to $x=5$. Further, notice (due to the right angles formed by a radius and its tangent line) that the quadrilateral (a kite) $AOBX$ is cyclic. Therefore, we can apply Ptolemy's Theorem to give: $2\sqrt{170}r=d\sqrt{40}$, where $r$ is the radius of the circle and $d$ is the distance between the circle's center and $(5,0)$. Therefore, $d=\sqrt{17}r$. Using the Pythagorean Theorem on the right triangle $OAX$ (or $OBX$), we find that $170+r^2=17r^2$, so $r^2=\frac{85}{8}$, and thus the area of the circle is $\boxed{\frac{85\pi }{8}}$.