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.
Best Answer
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}) $