That's a Circle, right?

We know $\mid 2x + 8 \mid$ is a piecewise function which equals to $2x + 8$ for $x \geq -4$ and $-2x - 4$ for $x < -4$

In this case, for $x \geq 4$ , it satisfy the equation $x^2 + y^2 - 2x - 8 + 2y = 16 \Rightarrow (x-1)^2 + (y+1)^2 = ( \sqrt {26} )^2$ , which is a circle with center $(1,-1)$ and radius $\sqrt{26}$

While for $x < 4$ , it satisfy the equation $x^2 + y^2 + 2x + 8 + 2y = 16 \Rightarrow (x+1)^2 + (y+1)^2 = ( \sqrt {10} )^2$ , which is a circle with center $(-1,-1)$ and radius $\sqrt{10}$

Solving them simultaneously, we get the intersection points $(-4, -2)$ and $(-4,0)$

Because the curvature of a circle is inversely proportional its radius, and $\frac {1}{\sqrt {10}} > \frac {1}{\sqrt {26}}$ , the graph that satisfy the given equation is a large circle satisfying the first equation but bends outwards to the left for $x < 4$ , this "bend" satisfy the second equation.

Thus, $A$ is the area of the large circle plus the area of the "bend part"

$\begin{aligned} A & = & \pi \cdot ( \sqrt {26} )^2 + \displaystyle \int_{-2}^0 \left ( \left ( 1 - \sqrt{26-(y+1)^2} \right ) - \left (- 1 - \sqrt{10-(y+1)^2} \right ) \right ) \mathrm{d}y \\ & = & 26 \pi + \displaystyle \int_{-2}^0 \left ( 2 - \sqrt{26-(y+1)^2} + \sqrt{10-(y+1)^2} \right ) \mathrm{d}y, \space \text{ let } z = y+1 \\ & = & 26 \pi + \displaystyle \int_{-1}^1 \left ( 2 - \sqrt{26-z^2} + \sqrt{10-z^2} \right ) \mathrm{d}z \\ & = & 26 \pi + 2 - 26 \sin^{-1} \left ( \frac {1}{\sqrt{26}} \right ) + 10 \sin^{-1} \left ( \frac {1}{\sqrt{10}} \right ) \\ & \approx & 81.76662998 \Rightarrow \lfloor 1000 \{ A \} \rfloor = \boxed{766} \\ \end{aligned}$

The figure is actually composed of two circles. When the modulus is removed, we get two circles as follows:

$C_1:x^2+y^2-2x+2y-24=0,$ $when$ $x \geq -4$

$C_2:x^2+y^2+2x+2y-8=0,$ $when$ $x < -4$

Area of a sector = $\pi R^2 \times \frac{\theta}{2 \pi} = \frac{\theta}{2} R^2$

For $C_1 ,$ $radius=AO=BO= \sqrt{26} ,$ $PO=5$ and $AP=PB=1$ .

Consider $\angle AOP= \angle POB = \alpha$ . So, $\alpha = tan^{-1} \frac{1}{5}$ .

Area of the yellow part $(A_1)$ = Area of major sector + Area of triangle = $(\pi - tan^{-1} \frac{1}{5})(26) + 2 \times \frac{1}{2} (5)(1)$

For $C_2 ,$ $radius=AO=BO= \sqrt{10} ,$ $PO=3$ and $AP=PB=1$ .

Consider $\angle AOP= \angle POB = \beta$ . So, $\beta = tan^{-1} \frac{1}{3}$ .

Area of the red part $(A_2)$ = Area of minor sector - Area of triangle = $(tan^{-1} \frac{1}{3})(10) - 2 \times \frac{1}{2} (3)(1)$ .

Thus, the area of the figure is $A=A_1 +A_2 = 81.76663$ and it's fractional part is $0.7666$ .

So, $100(0.7666)=766.6$ whose integral part is $\boxed{766}$ .

Apply the following formula for the area of a segment of a circle $Area = r^{2} \theta - r^{2} \sin \theta \cos\theta$ where $\theta$ is half the angle subtended by the chord on the center.

Now, the required area is nothing but area of the circle - the area of segment of the bigger circle + the area of segment of smaller circle.

Area comes out to be $\approx 81.766$ which is slightly more than $26 \pi$ Hence the required answer $\boxed{766}$