An amicable split .....

Let the origin $(0,0)$ be $O$ , the center of the circle $(0,a)$ be $P$ , the top of the circle $(0,a+2)$ be $Q$ , the line $y = x$ , $x \ge 0$ cuts the circle at $R$ and $\angle PQR = \theta$ .

Then, the area of the semicircle above the line $y = x$ , $x \ge 0$ = the area of the sector $PQR$ + the area of $\triangle OPR$ = $\dfrac {\pi r^2}{4}$ , where $r = 2$ is the radius of the circle.

Therefore, $\quad \dfrac {\theta}{2\pi} (\pi r^2) + \dfrac {1}{2} a \sin {\theta} = \dfrac {\pi r^2}{4}$ .

We note that $a = r \sin {\theta} - r \cos {\theta}$ .

$\Rightarrow \dfrac {\theta r^2}{2} + \dfrac {1}{2} r^2 \sin {\theta} (\sin {\theta} - \cos {\theta}) = \dfrac {\pi r^2}{4}$

$\Rightarrow \theta + \sin {\theta} (\sin {\theta} - \cos {\theta}) = \dfrac {\pi}{2}$

Using numerical method, we find that $\theta = 1.133617279$ rad.

Therefore, $a = 2 (\sin {\theta} - \cos {\theta}) = 2 (0.905949616 - 0.423385513)$

$= 2\times 0.482564104 = 0.965128208\quad \Rightarrow \lfloor 1000a \rfloor = \boxed {965}$

Just to get something posted, $a$ can be found from the equation

$2\arcsin((\frac{1}{4})(a + \sqrt{8 - a^{2}})) + (\frac{a}{4})(a + \sqrt{8 - a^{2}}) = \pi$ ,

yielding a value of $a = 0.9651285....$ , making $\lfloor 1000*a \rfloor = \boxed{965}$ .

I'll post how I reached this equation later, (no calculus involved), but what I'm hoping for is that someone can provide a solution, (geometric or calculus), in which the answer drops out more easily. It looks like a simple problem with a simple solution, but that's not what I've found it to be ......

EDIT; O.k., now for some details .....

Let the circle have equation $x^{2} + (y - a)^{2} = 4$ . For the area of the circle to be split as described, the region in the first quadrant between the circle and the line $y = x$ must have an area of $\pi$ .

Now we can split this region into two sections, namely (i) the sector of the circle between the $y$ -axis and the radius joining the center of the circle and the point $(X,Y)$ where the circle and $y = x$ intersect, and (ii) the triangle beneath this sector having base length $a$ and 'height' $X$ .

The point $(X,Y)$ is where $X^{2} + (X - a)^{2} = 4 \Longrightarrow 2X^{2} - 2aX + (a^{2} - 4) = 0$

$\Longrightarrow X = \dfrac{2a \pm \sqrt{4a^{2} - 8(a^{2} - 4)}}{4} = \dfrac{a \pm \sqrt{8 - a^{2}}}{2}$ .

Since we are looking here at the first quadrant, we take $X = \dfrac{a + \sqrt{8 - a^{2}}}{2}$ .

The central angle of the sector is $\theta = \arcsin \frac{X}{2}$ , so its area is $2*\arcsin \frac{X}{2}$ .

The area of the triangle below this sector is $(\frac{1}{2})aX$ .

The equation we thus need to solve for $a$ is thus

$2\arcsin((\frac{1}{4})(a + \sqrt{8 - a^{2}})) + (\frac{a}{4})(a + \sqrt{8 - a^{2}}) = \pi$ ,

yielding a value of $a = 0.9651285....$ , making $\lfloor 1000*a \rfloor = \boxed{965}$ .