The crucial observation is that the two equations imply $\sum_{i<j} (a_i-a_j)^2=9(a_1^2+\dots+a_9^2)-(a_1+\dots+a_9)^2=8b+6$ i.e. the quadratic terms cancel out. This means that all the differences between the $a_i$ are bounded by $\mathcal{O}(\sqrt{b})$. If we fix the six differences $a_2-a_1,\dots,a_7-a_1$, we thus have $\mathcal{O}(b^3)$ choices. It will thus suffice to prove that such a choice determines the nonuple up to $\mathcal{O}(b^{\varepsilon})$ many choices. Indeed, writing $a_1=a$ and $a_8=x, a_9=y$, we now have fixed $7a+x+y=C_1$ and $7a^2+x^2+y^2+C_2a=C_3$ where all the constants are bounded by polynomials in $b$. Solving the first equation for $a$ and plugging back into the second and simplifying we find that $(8x-C_4)^2+(8x-C_5)^2=C_6$ where $C_4,C_5,C_6$ are still bounded by polynomials in $b$. But now it follows from the classical theory of sums of two squares (e.g. via factorization over $\mathbb{Z}[i])$ that such an equation has at most $\mathcal{O}(C_6^{\varepsilon})$ and hence $\mathcal{O}(b^{\varepsilon})$ many solutions. Done!