jmc

Call the first number $x$ and the second number $y$. Without loss of generality, assume $x>y$. We can represent the information given in the problem with the following system of linear equations: $$\begin{array}{rl}x-y& =9\\ x^2+y^2& =153\end{array}$$ Solving for $x$ in the first equation and substituting into the second yields $(9+y{)}^2+y^2=153$, or $2y^2+18y-72=0$. Canceling a $2$ gives $y^2+9y-36=0$, which factors into $(y+12)(y-3)$. Thus, $y=3$ and $x=12$. So, $x\cdot y=\boxed{36}$.