I was working with two optimization problems and have a feeling that they are dual of each other.
First problem: we know that some convex orange set lies inside red unit circle and I want to find its maximum real part. We use the green line inside unit circle starting from $x=1$ and move it to the left until it touches the orange convex set. When we get the optimal green line, we calculate the angle $\alpha$, which is in the range $[0,\pi)$ if the orange set is not a point at $x=-1$, or $+\infty$ when the orange convex set is a point at $x=-1$.
Second problem: we know that some convex orange set lies inside red parabola $y=x^2$. We move green line $y=\frac{\lambda^2+1}{\lambda}x-1$, starting from $\lambda=1$ (which is equivalent to $\{x,y\}=\{1,1\}$) and until $\lambda=+\infty$ (which is equivalent to $\{x,y\}=\{0,+\infty\}$), and we want to move it until it touches the orange convex set.
I just guess that in the second figure, if we take a log scale on $x$ axis and then somehow connect two ends of red parabola, we will get the first figure. Then two problems are equivalent, in both cases the orange set cannot lie outside of unit circle, the range of $\alpha$ and $\log\lambda$ is from $0$ to $+\infty$, although I cannot explain why $\alpha$ is not continuous. and the search starts from $\alpha=0$ or $\log \lambda=0$ until the green line touches orange set.
How such transformation can be done? Is there any name?
Let's embed your parabola in three-dimensional space, that is, add a $z$-axis to the coordinate system. Construct a circle in the plane $z = -\sqrt3(y - 1)$ on a diameter with endpoints $A = \left(0,\frac12,\frac{\sqrt3}{2}\right)$ and $B = \left(0,\frac32,-\frac{\sqrt3}{2}\right).$ Construct a right circular cone on this circle, placing the vertex of the cone at $P = \left(0,-\frac12,-\frac{\sqrt3}{2}\right).$
In the figure below, all of these objects are shown in projection on the $y,z$ plane with the $y$-axis horizontal and the $z$-axis vertical. The circle is shown in profile as a diagonal broken red line, the cone lies between the orange lines at $y = 0$ and $z = -1,$ and the central projection through $P$ of the circle onto the $x,y$ plane is exactly where the cone intersects the plane, that is, it traces a parabola. Specifically, the projection is the parabola $y = x^2,$ shown in profile as a horizontal red ray along the $y$-axis.
The line through $P$ and $Q = (0,-1,0)$, shown in black in the figure, is given by the equations $z = -\sqrt3(y - 1)$ and $x = 0$, that is, it is parallel to $\overline{AB}.$ If you construct several additional planes through the line through $P$ and $Q$ so that each plane passes through the circle, these planes will intersect the circle in a set of parallel lines and will intersect the $x,y$ plane in a set of lines all passing through $Q$ (that is, the point on the $y$-axis at $y = -1$). Each of these lines in the $x,y$ plane is a line with an equation $$ y = \frac{\lambda^2 + 1}{\lambda} x - 1. $$
By a suitable change of coordinates, you can assign $x,y$ coordinates to the plane of the circle so that the center of the circle is at $(0,0)$ and the diameter $\overline{AB}$ lies on the $y$-axis. The parallel lines where planes through $\overline{PQ}$ intersect the circle all have equations of the form $x = \cos(\alpha)$ in these coordinates. But each of these lines is projected onto a line with an equation of the form $ y = \frac{\lambda^2 + 1}{\lambda} x - 1 $ in the original $x,y$ plane of the diagram above.
Since this is a central projection, any convex set in the plane of the circle is projected to a convex set in the plane of the parabola. In fact, convex polygons are projected onto convex polygons (in case this helps).
Geometrically the projection maps the two problems to each other perfectly. The only rough spot I can see is the form of the equation $ y = \frac{\lambda^2 + 1}{\lambda} x - 1. $ As $\lambda$ runs from $1$ to $+\infty,$ $\alpha$ runs from $0$ to $\frac\pi2.$ But the green line along the $y$-axis of the parabola exists only in the limit as $\lambda\to+\infty,$ and the green lines corresponding to $\frac\pi2<\alpha\leq\pi,$ that is, the ones on the left half of the parabola, can only be expressed by setting $\lambda < -1.$ So there is a discontinuity, and it's all due to the use of the parameter $\lambda.$
But what if instead of letting $\lambda$ run from $1$ to $+\infty,$ you let $\lambda$ run from $1$ to $0$? Once again the slope of the line increases from $2$ toward $+\infty.$ In the limit at $\lambda\to0$ the line becomes vertical, and as $\lambda$ becomes negative (decreasing toward $-1$) the line sweeps across the left half of the parabola. Now the only problem is the need to take a limit at $\lambda = 0$ since the equation of the line is then undefined.
We can fix this too by writing the equation of the line as $$ x = \frac{\lambda}{\lambda^2 + 1}(y + 1). $$ Then as $\lambda$ decreases from $1$ to $-1$ the green lines sweep across the parabola from right to left. You can make a one-to-one continuous mapping between the $\alpha$ parameter in the interval $[0,\pi]$ and the $\lambda$ parameter in the interval $[-1,1].$