Cartesian equation for a transcendental / trigonometric curve 
Hello! Please see figure above. I am searching for the cartesian equation for the curve in green, similar to how the equation for a semicircle is $f(x) = √(1 - x^2)$. I'm not sure if this is even possible, so any feedback is appreciated. My several attempts have failed.
The construction is set up like so (upon a circle of radius=1):
$$\frac{m}{k} = \frac{k}{t}$$
or
$$\frac{sin(α)} {sin(β)} = \frac{sin(β)} {sin(α+β)}$$
As you can see, $x$ is in blue and $f(x)$ is in green, and as α grows, then the point H traces out the curve in green, which I'm calling a gourd curve. Is it possible to obtain a cartesian equation for the gourd curve and not something which contains trigonometric identities? Trigonometric identities are also welcome, but I'd prefer to analyze the equation without them.
 A: Relabelling your figure a bit, I'm taking center $O$, and points $A$ and $B$ on the circle with $M$ the foot of the perpendicular from $B$ to $\overline{OA}$. If the circle meets the $x$-axis at $R$, then define $\alpha :=\angle ROA$ and $\beta :=\angle AOB$.

If the radius of the circle is $r$, then we can write the coordinates of $M$ as
$$x = r \cos\beta \cos\alpha \qquad y = r\cos\beta \sin\alpha \tag1$$
The imposted relation on $\alpha$ and $\beta$ is
$$\frac{\sin\alpha}{\sin\beta}=\frac{\sin\beta}{\sin(\alpha+\beta)} \tag2$$
Expanding $(2)$, and using the parameterizations
$$\cos\alpha=\dfrac{1-a^2}{1+a^2} \quad \sin\alpha=\dfrac{2a}{1+a^2} \quad
\cos\beta=\dfrac{1-b^2}{1+b^2} \quad \sin\beta=\dfrac{2b}{1+b^2} \tag3$$ we can use, say, the method of resultants to eliminate $a$ and $b$ to obtain this cartesian equation for the gourd curve:

$$\begin{align}
0 &= (x^2 + y^2)^8 \\
 &\quad- 2 r^2\, (x^2 + y^2)^5\, (2 x^4 + 3 x^2 y^2 + 3 y^4) \\
 &\quad+\phantom{2}r^4\, (x^2 + y^2)^4\, (6 x^4 + 6 x^2 y^2 + 11 y^4) \\
 &\quad- 2 r^6\, (x^2 + y^2)^2\, (2 x^6 + 3 x^4 y^2 + 5 x^2 y^4 + 3 y^6) \\
 &\quad+\phantom{2}r^8\, ((x^2+y^2)^2 - x^2 y^2)^2
\end{align} \tag{$\star$}$$

(This form was used to generate the figure above. Because $x$ and $y$ appear exclusively to even powers, the solution set includes a horizontally-mirrored copy of the gourd.)
In polar form (using $\rho$ for the radial variable since I'm using $r$ for the radius of the circle):

$$\begin{align}
0 &= \rho^8 \\
&\quad-2\rho^6 r^2\, (3 - 3 \cos^2\theta + 2 \cos^4\theta) \\
&\quad+\phantom{2}\rho^4 r^4\,  (11 - 16 \cos^2\theta + 11 \cos^4\theta) \\
&\quad-2 \rho^2 r^6\,  (3 - 4 \cos^2\theta + 2 \cos^4\theta + \cos^6\theta) \\
&\quad+\phantom{2\rho^2} r^8\, (1 - \cos^2\theta + \cos^4\theta)^2
\end{align} \tag{$\star\star$}$$

(Because $\rho$ appears exclusively to even powers, real roots occur in $\pm$ pairs; taking just one of these should give a single "gourd".)
