Limiting range

We note that $g(x)$ is a straight line with a gradient of $\frac{4}{3}$ and $f(x)$ is a dome-shape curve and a larger $a$ shifts $f(x)$ up and the largest $a$ is when $f(x)$ touches $g(x)$ . That is when $f(x) = g(x)$ and $\frac{d}{dx}f(x) = \frac{d}{dx}g(x) = \frac{4}{3}$ .

$\dfrac{d}{dx}f(x) = \dfrac{1}{2}\left( \dfrac {-2x-2}{\sqrt{-x^2-4x}} \right) = - \dfrac {x+4}{\sqrt{-x^2-4x}}$

$\begin{aligned} \Rightarrow - \dfrac {x+2}{\sqrt{-x^2-4x}} & = \dfrac {4}{3} \\ -3(x+2) & = 4 \sqrt{-x^2-4x} \\ 9(x^2+4x+4) & = -16(x^2+4x) \\ 25x^2 + 100x + 36 & = 0 \\ (5x-2)(5x-18) & = 0 \end{aligned}$

$\Rightarrow x = \begin{cases} - \dfrac{2}{5} & \Rightarrow a + \sqrt{-(- \frac{2}{5})^2 -4(- \frac{2}{5})} \le 1 + \dfrac{4(- \frac{2}{5})}{3} \\ & \quad \space a + \sqrt{\frac{144}{100}} \le \dfrac {7}{15} \quad \Rightarrow a \le - \dfrac{11}{15} \\ - \dfrac{18}{5} & \Rightarrow a + \sqrt{-(- \frac{18}{5})^2 -4(- \frac{18}{5})} \le 1 + \dfrac{4(- \frac{18}{5})}{3} \\ & \quad \space a + \sqrt{\frac{144}{100}} \le -\dfrac {19}{5} \quad \Rightarrow a \le - 5 \end{cases}$

Therefore, the biggest $a$ is $\boxed{-5}$ .