Given a circle, and a set of points $A$ that lie external to the circle; I perform the following simple operation:
I compute the point of intersection of the i) circle and the ii) line joining each external point and the center of the circle. This gives a set of points $T$ which lie on the circle.
Question: Now would the rank ordering of the 'Euclidean distances' between the points in $A$ be preserved by the rank orderings of the corresponding 'Circular distances' between the points in $T$?
No, it's not preserved. Think of the points as vectors from the origin. Then the square of the Euclidean distance between $u$ and $v$ is $$|u-v|^2=|u|^2+|v|^2-2|u||v|\cos \theta\tag1$$ where $\theta $ is the corresponding circular distance, $0\le \theta\le \pi$.
Now, $-\cos \theta$ is indeed an increasing function of $\theta$ on the interval $[0,\pi]$. However, the magnitudes $|u|$ and $|v|$ also appear on the right side of (1). Pairs of vectors with the same value of $\theta$ can have quite different Euclidean distances $|u-v|$.
For a concrete example, take $A=(3,0)$, $B=(10,1)$ and $C=(3,1)$. Clearly $|AB|>|AC|$. Yet, the projected points $\tilde A=(1,0)$, $\tilde B=(10/\sqrt{101},1/\sqrt{101})$ and $\tilde C=(3/\sqrt{10},1/\sqrt{10})$ satisfy the opposite $|\tilde A\tilde B|<|\tilde A\tilde C|$.