Ans: A
Note that $\Delta OAB$ is an isosceles triangle with $OA=OB$.
Note that $\Delta OAB$ is an isosceles triangle with $OA=OB$.
Therefore the orthocentre must lie on the $x$-axis. Let $G(x,0)$ be the orthocentre.
$\begin{array}{rcl}
m_{AG} \times m_{OB} & = & -1 \\
\dfrac{-24-0}{18-x}\times \dfrac{24-0}{18-0} & = & -1 \\
18(18-x) & = & 576 \\
18-x & = & 32 \\
x & = & -14
\end{array}$
