Geometry problem using circular arcs

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The picture above illustrates how $A_1 + A_2$ depends on the position of $T$. For every $T$ there is a point on the vertical axis with the same sum of the arcs. Therefore we can consider a unit circle with center $C = (0,c)$ that intersects the horizontal axis at $(0,\pm x)$ where $x=\sqrt{1-c^2}$ and place $T$ at $(0,y)$

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The length of the arcs is then given by $A = 2\arccos{(x)} + \pi$ and $A_1 = A_2 = r \cdot \alpha$ where

$$\begin{matrix} \displaystyle r = \sqrt{a^2 + \left(y - c \left(\frac{a}{x} + 1 \right) \right)^2} \\ \displaystyle \alpha = \arccos{\left( \frac{a + x}{1 - c^2} \, \frac{a + cx(c-y)}{r^2} \right)} \end{matrix} \qquad a = \frac{x}{2} \, \frac{(c-y)^2 - 1}{1 + c(y-c)}$$

So your problem becomes just algebra $$\arccos{(x)} + \frac{\pi}{2} > r \cdot \begin{cases} \alpha \\ (2 \pi - \alpha)\end{cases} \begin{matrix} \text{if} \; 0 < y - c < 1 \\ \text{if} \; 0 < y < c \end{matrix}$$