A concentric circles related problem. Given a circle of radius $AC= a$ with center in $C(C_x,0)$ and with $C_x>0$. Given an angle $\beta$ (between the points B C F) and a smaller circle of radius $BC =\displaystyle \frac{a+b}{2}$ , also with center in $C$ where the measure $\displaystyle b = \frac{a(1-\sin \beta)}{1+\sin \beta}$. What measure must the angle $\beta$ have  for the condition $\displaystyle \left(\frac{a+b}{2}\right)\cos \beta = C_x$ to be true? Which should also mean that the point $\displaystyle F\left(\frac{(a+b)}{2}\sin \beta,\; \frac{(a+b)}{2}\cos \beta\right)$ belongs to the $y$-axis.

My attempt so far was to replace $b$ with
$\frac{a(1-\sin \beta)}{1+\sin \beta}$
in $\displaystyle \left(\frac{a+b}{2}\right)\cos \beta = C_x$
and trying to solve
$\displaystyle \left(\frac{a+\frac{a(1-\sin \beta)}{1+\sin \beta}}{2}\right)\cos \beta = C_x$
then I replaced $cos \beta$ with the sqrt of $1-sin^2\beta$
$\Biggl(\displaystyle \left(\frac{a+\frac{a(1-\sin \beta)}{1+\sin \beta}}{2}\right)\Biggl)^2(1-sin^2\beta)= C^2_x$
$\displaystyle \frac{a^2}{4}\left(1+\frac{1-sin\beta}{1+sin\beta}\right)^2\left(1-sin^2\beta\right)= C^2_x$
but at this point I got lost.
I'm wondering if I am missing a much easier solution or if I am going completely the wrong way.
UPDATE: I made a mistake in setting up the problem this morning when I posted the question, specifically in setting the value of BC. My most sincere apologies, it is now corrected.
 A: Thanks to @Aretino and @SarGe for the guidance (and patience).
$\displaystyle \frac{a^2}{4}\left(1+\frac{1-sin\beta}{1+sin\beta}\right)^2\left(1-sin^2\beta\right)= C^2_x$  gives two solutions
$sin\beta= \frac{-C^2_x \pm\sqrt{C^4_x-(C^2_x+a^2)(C^2_x-a^2)}}{C^2_x+a^2} $
After testing, the correct solution resulted to be
$sin\beta= \frac{-C^2_x +\sqrt{C^4_x-(C^2_x+a^2)(C^2_x-a^2)}}{C^2_x+a^2} $
So
$\beta = asin(\frac{-C^2_x +\sqrt{C^4_x-(C^2_x+a^2)(C^2_x-a^2)}}{C^2_x+a^2}) $
A: Let's define $\theta:=\frac12(90^\circ-\beta)$, so that $\beta=90^\circ-2\theta$, as this gives
$$b = a\frac{1-\sin\beta}{1+\sin\beta}=a\frac{1-\cos2\theta}{1+\cos2\theta}=a\frac{2\sin^2\theta}{2\cos^2\theta}=a\tan^2\theta \tag{1}$$
Then, defining $c:=C_x$ to reduce visual clutter, the target condition becomes
$$c=\frac12(a+b)\cos\beta=\frac{a}2(1+\tan^2\theta)\cdot\sin2\theta=\frac{a}{2}\cdot\frac{1}{\cos^2\theta}\cdot2\cos\theta\sin\theta=a\tan\theta \tag{2}$$
Therefore,

$$\theta=\arctan\frac{c}{a} \quad\to\quad \beta = 90^\circ-2\arctan\frac{c}{a} \tag{$\star$}$$

To see that this is consistent with @OP's answer, first note that OP's answer reduces to
$$\sin\beta = \frac{-c^2\pm \sqrt{c^4-(c^2+a^2)(c^2-a^2)}}{a^2+c^2}=\frac{-c^2\pm \sqrt{a^4}}{a^2+c^2}\quad\stackrel{\sin\beta\,\geq\,0}{\to}\quad\sin\beta=\frac{a^2-c^2}{a^2+c^2} \tag{3}$$
On the other hand, from $(\star)$,
$$\sin\beta=\cos\left(2\arctan\frac{c}{a}\right)=\cos^2\arctan\frac{c}{a}-\sin^2\arctan\frac{c}{a}=\frac{a^2}{a^2+c^2}-\frac{c^2}{a^2+c^2}\tag{4}$$
which matches $(3)$. $\square$

To address a refinement in the comments, we replace $(1)$ by
$$b=a\frac{1-\sin\beta_n}{1+\sin\beta_n} \qquad \beta_n := \frac{\beta}{n} \tag{1'}$$
for odd integer $n$. We can again define $\theta := \tfrac12(90^\circ-\beta_n)$ so that, as before, $$b = a \tan^2\theta$$
Things get more complicated with $(2)$, which we can start to write as
$$c = \frac12(a+b)\cos(n\beta_n)=\frac12a\sec^2\theta\sin(90^\circ n)\sin 2n\theta \tag{2'}$$
Of course, $\sin(90^\circ n)=\pm 1$ for odd $n$, so that's not a problem; however, expanding $\sin2n\theta$ quickly becomes one, even for $n=3$ and $n=5$. Abbreviating $\tan\theta$ with $t$, we have  ...

*

*$n=3$:
$$\begin{align}
2c\cos^2\theta &= -a\sin 6\theta \\[4pt]
\to\quad2c\cos^2\theta &= - 2a\sin\theta\cos\theta\left( 4 \cos^2\theta - 1\right) \left( 4 \cos^2\theta - 3 \right) \\[4pt]
\to\quad\phantom{2\cos^2\theta} c &= - at \left(\frac{4}{1+t^2}-1\right)\left(\frac{4}{1+t^2}-3\right) \\[4pt]
\to\quad\phantom{c\cos^2\theta} 0 &= 3 a t^5 + c t^4 - 10 a t^3 + 2 c t^2 + 3 a t + c \tag{2'.3}   
\end{align}$$


*$n=5$:
$$\begin{align}
2c\cos^2\theta &= a\sin 10\theta \\[4pt]
\to\quad\phantom{c\cos^2\theta} 0 &= 5at^9-ct^8-60at^7-4ct^6+126at^5 \\
&\phantom{=\;}- 6ct^4 - 60at^3 - 4ct^2 + 5at - c \tag{2'.5}   
\end{align}$$
We almost-never get explicit formulas for $t=\tan\theta$ satisfying $(2'.3)$ and $(2'.5)$. Numerical methods will be required. Once appropriate values of $\theta$ are found, however, we get $\beta$ via
$$\beta = n\beta_n = n (90^\circ-2\theta) \tag{$\star$'}$$
