Points $A=(6,13)$ and $B=(12,11)$ lie on circle $\omega$ in the plane. Suppose that the tangent lines to $\omega$ at $A$ and $B$ intersect at a point on the $x$-axis. What is the area of $\omega$?
$\textbf{(A) }\frac{83\pi}{8}\qquad\textbf{(B) }\frac{21\pi}{2}\qquad\textbf{(C) } \frac{85\pi}{8}\qquad\textbf{(D) }\frac{43\pi}{4}\qquad\textbf{(E) }\frac{87\pi}{8}$
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](https://artofproblemsolving.com/wiki/index.php/Ptolemy%27s_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 $\boxed{\textbf{(C) }\frac{85}{8}\pi}$.

Diagram for Solution 1
[caption](https://artofproblemsolving.com/wiki/index.php/File:Desmos-graph_(1).png)
~BakedPotato66