Geometry problem: Line intersecting  a semicircle Suppose we have a semicircle that rests on the negative x-axis and is tangent to the y-axis.A line intersects both axes and the semicircle. Suppose that the points of intersection create three segments of equal length. What is the slope of the line?
I have tried numerous tricks, none of which work sadly.

Let the line meet the axes at $A(-a,0)$ and $B(0,b)$, and the semi-circle at $A^\prime$ and $B^\prime$ (with $A^\prime$ the closer to $A$, and $B^\prime$ the closer to $B$). Let $O$ be the origin, and define $d := |AA^\prime|=|A^\prime B^\prime|=|B^\prime B| > 0$ as the common length of the segments.
The Power of a Point Theorem, applied to point $B$, tells us that 
$$\begin{eqnarray*}
|BB^\prime|\cdot|BA^\prime| &=& |BO|^2 \\
\implies d \cdot (2d) &=& b^2 \\
\implies 2 d^2 &=& b^2
\end{eqnarray*}$$
Also, Pythagoras tells us that
$$\begin{eqnarray*}
a^2 + b^2 = ( 3 d )^2 = 9 d^2
\end{eqnarray*}$$
Eliminating $b$, we have
$$a^2 = 7 d^2$$
so that the slope is
$$\frac{b}{a} = \frac{\sqrt{2}d}{\sqrt{7}d}=\sqrt{\frac{2}{7}}$$
NOTE. If we cared for the actual value of $d$, we could leverage the Power of Point $A$ (writing $P$ for the point $(-2,0)$):
$$\begin{eqnarray*}
|AP|\cdot|AO| &=& |AA^\prime|\cdot|AB^\prime|\\
\implies (2-a)\cdot a &=& 2 d^2 \\
\implies (2-d\sqrt{7})\cdot d\sqrt{7} &=& 2 d^2 \\
\implies \frac{2\sqrt{7}}{9} &=& d
\end{eqnarray*}$$
