Points $A(4,10)$ and $B(10,2)$ 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 $\omega$?

1. Let $C$ be the intersection of the tangents at $A,B$; the midpoint of $AB$ is $M=(7,6)$, and line $CM$ (perpendicular to $AB$, since $AB$ has slope $-\frac43$ so $CM$ has slope $\frac34$) meets the $x$-axis at $C=(-1,0)$.
2. $CA^2=(4-(-1))^2+10^2=125$, $CM^2=8^2+6^2=100$. By theorem 7, $CO=\dfrac{CA^2}{CM}=\dfrac{125}{10}=12.5$.
3. $OA^2=CO^2-CA^2=156.25-125=31.25=\dfrac{125}4$.
4. The answer is $\boxed{\dfrac{125\pi}4}$.