Then let us say that the arc has length \(\pi - 2\alpha\) and is placed symmetrically around the \(y\)-axis as in this diagram. Then the arc itself is the segment of the circle between the points \((\cos \alpha,\sin \alpha)\) and \((-\cos \alpha,\sin \alpha)\) in the upper half of \(\mathbb{R}^2.\) And its chord is the horizontal line segment connecting its endpoints as shown.
If we take a random point \(p=(\sin t,\cos t)\) in the upper right quadrant, i.e. the point for which its distance to the \(y\)-axis along the arc is \(t,\) with \(0 < t < \pi/2,\) then the line defined by this point is the line \(L\) through \(p\) and the point \(q=(s,\sin \alpha)\) on the chord, such that the ratio \(s/\cos \alpha\) of the distance from the \(y\)-axis to \(q\) equals the ratio \(t/(\pi/2-\alpha)\) of the distance along the arc between the \(y\)-axis and \(p\) itself to the entire piece of arc in the upper right quadrant. So \(s = \frac{\cos \alpha}{\pi/2-\alpha}t\) follows.
The equation for \(L\) becomes \(L : y = \frac{\cos t - \sin \alpha}{\sin t -s}(x - s) + \sin \alpha,\) which is checked by setting \(x = s,\sin t.\)
Clearly the \(y\)-intercept of \(L\) is the point \((0,-\frac{\cos t - \sin \alpha}{\sin t - s}s+\sin \alpha)\) by setting \(x=0\) in the equation for \(L.\)
If this is correct, then the question seems to be whether there exists a value of \(\alpha\) with \(0 < \alpha < \pi/2\) for which this intersection point does not depend on \(t.\)