Supporting the $\sin$

By symmetry, let equations be

${ \left( x-\frac { \pi }{ 2 } \right) }^{ 2 }+{ \left( y-c \right) }^{ 2 }={ r }^{ 2 }\quad and\quad { \left( x+\frac { \pi }{ 2 } \right) }^{ 2 }+{ \left( y-d \right) }^{ 2 }=4{ r }^{ 2 }$

Distance between centers=3r

$\Rightarrow { \pi }^{ 2 }+{ \left( c-d \right) }^{ 2 }=9{ r }^{ 2 }$

Slope of radicle axis of circles=slope of tangent at (a,sin(a))

$\Rightarrow cos(a)=\frac { \pi }{ d-c }$

Shortest distance of tangent at (a,sin(a)) from center=radius of circle with that center

$\frac { |\frac { \pi }{ 2 } cos(a)+sec(a)\left( \frac { \pi }{ 2 } -a \right) -acos(a)| }{ \sqrt { { cos }^{ 2 }(a)+1 } } =\frac { \pi }{ 3 } \sqrt { { sec }^{ 2 }(a)+1 }$

Its easy to find b=1/2 and a= $\frac{ \pi }{ 6 }$ and by symmetry, b=-1/2 also.

For so long there hasn't been a solver.

Anyway, here's the solution:

So if a circle is tangent to a line, the normal of that line (at the point of tangency) would pass right through the center of the circle. So to find the answer it is obvious that we have to find the gradient of $y=\sin{(x)}$ .

The plan is to find the radius of the circle in terms of $b$ .

Consider this hi

$z=\pi -2\arcsin { (b) }$ $k=\pi +2\arcsin { (b) }$

Now lets focus on the circle fitting inside $z$ , with radius $R$

Gradient at $x=\arcsin{(b)}$ is $\cos{(\arcsin{(b)})}$

hi

Using the gradient above you can find that $\theta =\frac { \pi }{ 2 } -\arctan { (\cos { (\arcsin { (b))) } } }$

And with some more simple trigonometry you can finally express $R$ in terms of $z$

$R=\frac { z }{ 2\cos { (\theta ) } } =\frac { \pi -2\arcsin { (b) } }{ 2\cos { \left( \frac { \pi }{ 2 } -\arctan { \left( \cos { \left( \arcsin { (b) } \right) } \right) } \right) } }$

Now, doing the same on the circle fitting inside $k$ , with radius $r$ , we would find that

$r=\frac { k }{ 2\cos { (\theta ) } } =\frac { \pi +2\arcsin { (b) } }{ 2\cos { \left( \frac { \pi }{ 2 } -\arctan { \left( \cos { \left( \arcsin { (b) } \right) } \right) } \right) } }$

The question asks for $2$ circles with area ratio $\frac{1}{4}$ , so the radius ratio would be $\frac{1}{2}$ . So to find $b$ :

$r=2R$ $\frac { \pi +2\arcsin { (b) } }{ 2\cos { \left( \frac { \pi }{ 2 } -\arctan { \left( \cos { \left( \arcsin { (b) } \right) } \right) } \right) } } =\frac { \pi -2\arcsin { (b) } }{ \cos { \left( \frac { \pi }{ 2 } -\arctan { \left( \cos { \left( \arcsin { (b) } \right) } \right) } \right) } }$ $\pi +2\arcsin { (b) } =2\left( \pi -2\arcsin { (b) } \right)$ $b=\frac{1}{2}$

Note that $b$ can also be $-\frac{1}{2}$ which settles ${b}_{1}$ and ${b}_{2}$

Plugging $b=\frac{1}{2}$ to find $r$ and $R$ to find the answer would give

$\left\lfloor 10000\left( A+ \left( {b}_{1} - {b}_{2}\right)\right) \right\rfloor = \boxed{411933}$

This is probably the craziest solution I wrote, I took a whole full hour.