Any way you slice it

The $3$ subregions will each have an area of $\frac{\pi}{6}$ .

For the "upper" subregion, let $2\theta$ be the angle subtended by the chord formed by the intersection of the circle with the line of greater slope. Then the area of this subregion is the area of the sector defined by the angle $2\theta$ minus the area of the triangle with the endpoints of the chord and the point $(1,0)$ as vertices. This gives us the equation

$\theta - \sin(\theta)\cos(\theta) = \frac{\pi}{6}$ ,

which, by way of numerical methods, has solution $\theta = 0.984484...$ radians. The angle that the line makes with the $x$ -axis is then $\frac{\pi}{2} - \theta$ , and so the slope of the line is $\tan(\frac{\pi}{2} - \theta) = 0.6642279...$ .

For the "lower" subregion, let $\alpha$ be the angle between the $x$ -axis and the radial line to the (non-origin) endpoint of the chord formed by the intersection of the circle with the line of least slope. Then the area of this subregion will be the sum of the areas of the sector defined by $\alpha$ and the triangle with the endpoints of the chord and the point $(1,0)$ as vertices. This gives us the equation

$\frac{\alpha}{2} + \sin(\frac{\pi - \alpha}{2})\cos(\frac{\pi - \alpha}{2}) = \frac{\pi}{6} \Longrightarrow \alpha + \sin(\alpha) = \frac{\pi}{3}$ ,

which, by way of numerical methods, has solution $\alpha = 0.536267...$ . The angle that the line makes with the $x$ -axis is then $\frac{\pi}{2} - \frac{\pi - \alpha}{2} = \frac{\alpha}{2}$ , and so the slope of the line is $\tan(\frac{\alpha}{2}) = 0.2747497....$ .

The sum of the two slopes is then

$S = 0.6642279... + 0.2747497.... = 0.9389776...$ ,

and so $\lceil 1000*S \rceil = \boxed{939}$ .

Here is a diagram of the solution lines. (Ignore the lower part of the circle.)

Converting the curve into polar form, we have:

$(x-1)^2+y^2=1 \\ x^2+y^2-2x=0 \\ r^2-2r \cos\theta=0 \\ (r-\cos\theta)^2-\cos^2\theta=0 \\ r=\sqrt{\cos^2\theta}+\cos\theta \\ r=2\cos\theta$

Let the equation of the $\color{#D61F06}{\text{red}}$ line be $y={k}_{1} x$ , and the equation of the $\color{#3D99F6}{\text{blue}}$ line be $y={k}_{2} x$ . These lines represent the lines $\theta = \tan^{-1} {k}_{1}$ and $\theta = \tan^{-1} {k}_{2}$ on the polar graph respectively. We now need the following to be true:

$\cfrac { 1 }{ 2 } \int _{ 0 }^{ \tan ^{ -1 }{ { k }_{ 1 } } }{ { \left( 2\cos { \theta } \right) }^{ 2 } } \, d\theta \, = \, \cfrac { \pi }{ 3 } \quad \text{and} \quad \cfrac { 1 }{ 2 } \int _{ 0 }^{ \tan ^{ -1 }{ { k }_{ 2 } } }{ { \left( 2\cos { \theta } \right) }^{ 2 } } \, d\theta \, = \, \cfrac { \pi }{ 6 }$

Firstly,

$\cfrac { 1 }{ 2 } \int { { \left( 2\cos { \theta } \right) }^{ 2 } } \, d\theta \\ = 2\int { \cos ^{ 2 }{ \theta } } \, d\theta \\ = \int { \cos { 2\theta } +1 } \, d\theta \\ = \cfrac { 1 }{ 2 } \sin { 2\theta } +\theta +C$

Therefore

$\left[ \cfrac { 1 }{ 2 } \sin { 2\theta } +\theta \right] _{ 0 }^{ \tan ^{ -1 }{ { k }_{ 1 } } }=\cfrac { \pi }{ 3 } \quad \text{and} \quad \left[ \cfrac { 1 }{ 2 } \sin { 2\theta } +\theta \right] _{ 0 }^{ \tan ^{ -1 }{ { k }_{ 2 } } }=\cfrac { \pi }{ 6 }$

This gives us ${k}_{1}=0.66422 \dots$ and ${k}_{2}=0.27475 \dots$

Therefore $S=0.93897 \dots$ and $\left\lceil 1000S \right\rceil =\left\lceil 1000(0.93897 \dots) \right\rceil =\left\lceil 938.97 \dots \right\rceil =\large \color{#20A900}{\boxed{939}}$ .

The strategy is to solve in rotated co-ordinates.This makes the problem simpler and easy to visualize.

My solution is identical to the Brian.

Since there has been the best way provided by others, I would only tell you another way but not in detail about its every single step. However, you can see the consequence and get some idea about what happens.

Most immediate way may not be an ultimate choice, but it can be fastest in a situation. I applied integration. Since iteration method is expected, type of approach applied is not critical.

$y = 1 - (x - 1)^2$

$\Rightarrow y = \sqrt{x (2 - x)}$

To intercept with $y = m x$ ,

$(m x)^2 = (\sqrt{x (2 - x)})^2$ .

$x = \frac{2}{1 + m^2}$ and $y = \frac{2 m}{1 + m^2}$

Therefore, $(\frac{2}{1 + m_1^2}, \frac{2 m_1}{1 + m_1^2})$ and $(\frac{2}{1 + m_2^2}, \frac{2 m_2}{1 + m_2^2})$ are the points of lines intercept with the semicircle.

My convention here is following the sequence of angle about positive x-axis for anti-clockwise of 1, 2 and 3!

Let $A_1 = \frac{2 m_1}{(1 + m_1^2)^2}$ + $\displaystyle \int_{\frac{2}{1 + m_1^2}}^2 \sqrt{x (2 - x)} d x$

$A_3 = \displaystyle \int_0^{\frac{2}{1 + m_2^2}}\sqrt{x (2 - x)} d x$ - $\frac{2 m_2}{(1 + m_2^2)^2}$

$A_2 = \frac{2 m_2}{(1 + m_2^2)^2}$ + $\displaystyle \int_{\frac{2}{1 + m_2^2}}^{\frac{2}{1 + m_1^2}}\sqrt{x (2 - x)} d x$ - $\frac{2 m_1}{(1 + m_1^2)^2}$ {For purpose to check.}

$A_1 = A_2 = A_3 = \frac{\pi}{6}$ since $\pi 1^2 \div 6 = \frac{\pi}{6}$ for radius of 1.

Let x = $2 \sin^2 \theta$ , $d x$ = 4 $\sin \theta \cos \theta d\theta$ and $\theta = \sin^{-1} \sqrt{\frac{x}{2}}$ :

$\int \sqrt{x ( 2 - x)} d x$ = $\theta - \frac14 \sin 4 \theta$ + C

$A_1 = \frac{\pi}{6} = \frac{2 m_1}{(1 + m_1^2)^2}$ + $\frac{\pi}{2} - \sin^{-1}\frac{1}{\sqrt{1 + m_1^2}} + \frac14 \sin (4 \sin^{-1}\frac{1}{\sqrt{1 + m_1^2}})$

$\Rightarrow \sin^{-1}\frac{1}{\sqrt{1 + m_1^2}} - \frac14 \sin (4 \sin^{-1}\frac{1}{\sqrt{1 + m_1^2}})$ - $\frac{2 m_1}{(1 + m_1^2)^2} = \frac{\pi}{2} - \frac{\pi}{6} = g (m_1) = \frac{\pi}{3}$

$A_3 = \sin^{-1}\frac{1}{\sqrt{1 + m_2^2}} - \frac14 \sin (4 \sin^{-1}\frac{1}{\sqrt{1 + m_2^2}})$ - $\frac{2 m_2}{(1 + m_2^2)^2} = g (m_2)= \frac{\pi}{6}$

Here we are, $g(m_1) = \frac{\pi}{3}$ while $g(m_2) = \frac{\pi}{6}$ for same function!

By iteration where only one value for each, $m_1$ and $m_2$ are obtainable $directly$ with Excel:

1
2
3
4

0.274749695046391   1.04719755119660
0.664227387875901   0.523598775598299
0.938977082922292   
939 

$m_1 =$ 0.27474969504639+

$m_2 =$ 0.66422738787590+

$A_2 = \frac{2 m_2}{(1 + m_2^2)^2} + \sin^{-1}\frac{1}{\sqrt{1 + m_1^2}} - \frac14 \sin (4 \sin^{-1}\frac{1}{\sqrt{1 + m_1^2}}) - \sin^{-1}\frac{1}{\sqrt{1 + m_2^2}} + \frac14 \sin (4 \sin^{-1}\frac{1}{\sqrt{1 + m_2^2}}) - \frac{2 m_1}{(1 + m_1^2)^2} = \frac{\pi}{6}$ for checking purpose. Verified to match for being exact value of $\frac{\pi}{6} =$ 0.523598775598299 as proximity to confirm correct.

As I like to find iterated values to high accuracy, my figures are very accurate.

$\lceil 1000 \times 0.938977082922292 \rceil = 939$

Answer: $\boxed{939}$