Why does the circle intersect the line? 
Is there an easy geometric way to prove that the circle is tangent to the line $\overleftrightarrow {CD}$, where $C=(3,-6)$, $D=(6,-2)$, and $B=(6,0)$? I can do this by using calculus but I think there has to be a nicer/shorter solution.
Thanks in advance.
Edit: I found that $CE$ has distance $3$, so maybe one could use the theorem of Thales to obtain a right angle in the triangle $(0,-6), E,(6,-6)$ and use this?

Yes, you set the equation for the circle equal to the equation of the line and solve for x. If there is only one solution, the line is tangent and if two solutions the line cuts through the circle.
$\sqrt{6^2 - x^2} = \frac{4}{3}x - 10$
$36 - x^2 = \frac{16}{9}x^2 - \frac{80}{3}x + 100$
$\frac{25}{9}x^2 - \frac{80}{3}x + 64 = 0$
$(\frac{5}{3}x - 8)^2 = 0$
$\frac{5}{3}x = 8$
$x = 4.8$
