Is it a segment?

First, to find the $x$ -coordinate that maximizes the length of $AB$ , we need to find which value of $x$ maximizes the length function $f(x) = \sqrt{1 - x^{2}} - (x + 1),$ (as the circle has equation $x^{2} + y^{2} = 1$ ). Then

$f'(x) = \dfrac{-x}{\sqrt{1 - x^{2}}} - 1 = 0$ when

$-x = \sqrt{1 - x^{2}} \Longrightarrow x^{2} = 1 - x^{2} \Longrightarrow x = -\dfrac{1}{\sqrt{2}},$

as we are clearly looking for a negative $x$ -value. Now $f''(x) = -\dfrac{1}{(1 - x^{2})^{\frac{3}{2}}} \lt 0$ for $x = -\dfrac{1}{\sqrt{2}}$ so by the second derivative test we can conclude that the length of $AB$ is maximized for this value of $x.$

Now define the points $P(-1,0), Q(-\frac{1}{\sqrt{2}})$ and $O(0,0)$ . Then the area of the shaded region will be the area of sector $AOP$ minus the combined areas of $\Delta AOQ$ and $\Delta BPQ.$ Now as $\dfrac{1}{\sqrt{2}} = \cos(\dfrac{\pi}{4})$ we know that $\angle AOP = \dfrac{\pi}{4},$ so the area of sector $AOP$ is just $\dfrac{\pi}{4}.$

Next, as $\Delta AOQ$ and $\Delta BPQ$ are both isosceles right triangles, their combined areas will be

$\dfrac{1}{2}*\left(\dfrac{1}{\sqrt{2}}\right)^{2} + \dfrac{1}{2}\left(1 - \dfrac{1}{\sqrt{2}}\right)^{2} = \dfrac{1}{4} + \dfrac{1}{2}*\left(\dfrac{3}{2} - \sqrt{2}\right) = 1 - \dfrac{\sqrt{2}}{2}.$

Thus the area of the shaded region is $A = \dfrac{\pi}{8} - 1 + \dfrac{\sqrt{2}}{2} = 0.09980586....$ and so $\lfloor 100A \rfloor = \boxed{9}.$