Because these two points are apart of the diameter, their midpoint will be the radius. You can create the equation $(x-\frac{3\sqrt{7}}{2})^{2}+(y-\frac{7\sqrt{3}}{2})^{2}=52.5$ Plug in $y=0$ to get your answer.

[asy][asy]xaxis();yaxis(); draw(Circle((3sqrt(7)/2,7sqrt(3)/2),sqrt(63+49*3)/2)); dot((0,0));dot((3sqrt(7),7sqrt(3)));dot((3sqrt(7),0)); draw((0,0)--(3sqrt(7),7sqrt(3))--(3sqrt(7),0)--cycle);[/asy][/asy] Let $O(0,0)$ and $A(3\sqrt7,7\sqrt3)$, and the other $x$ intercept be $P$. Note that the three points $OAP$ form a right triangle. Therefore, $OP\perp AP$. $AP$ is perpendicular to the $x$-axis. The answer is $\boxed{(3\sqrt7,0)}$.