Assume that the circle mentioned in the problem is the unit circle on the complex plane. Then we can say every vertex $A_i$ is equivalent to a complex number $z_i$ such that $|z_i| = 1$. On the other hand we have $d(A_i, A_j) = |z_i - z_j|$. So we should prove that
$$\sum_{1 \le i < j \le k} |z_i - z_j|^2 \le k^2.$$
We know that
$$|z_i - z_j|^2 = (z_i - z_j)(\bar{z}_i - \bar{z}_j) = |z_i|^2 + |z_j|^2 - (\bar{z}_i z_j + \bar{z}_j z_i) = 2 - \bar{z}_i z_j - \bar{z}_j z_i.$$
So we have
$$\sum_{1 \le i < j \le k} |z_i - z_j|^2 = 2 \binom{k}{2} - \sum_{1 \le i < j \le k} (\bar{z}_i z_j + \bar{z}_j z_i).$$
and
$$\begin{align*} \sum_{1 \le i < j \le k} (\bar{z}_i z_j + \bar{z}_j z_i) &= \sum_{1 \le i \ne j \le k} \bar{z}_i z_j = \sum_{1 \le i \le k} \bar{z}_i (z_1 + \dots + z_k - z_i) \\ &= \left( \sum_{1 \le i \le k} \bar{z}_i \right) \left( \sum_{1 \le i \le k} z_i \right) - k = \left| \sum_{1 \le i \le k} z_i \right|^2 - k. \end{align*}$$
Therefore,
$$\sum_{1 \le i < j \le k} |z_i - z_j|^2 = k^2 - k - \left( \left| \sum_{1 \le i \le k} z_i \right|^2 - k \right) = k^2 - \left| \sum_{1 \le i \le k} z_i \right|^2.$$
Since $\left| \sum_{1 \le i \le k} z_i \right|^2 \ge 0$, we have
$$\sum_{1 \le i < j \le k} |z_i - z_j|^2 \le k^2.$$
Equality holds whenever $\sum_{1 \le i \le k} z_i = 0$, in other words, when the circumcenter and the centroid of $A_1A_2\dots A_k$ coincide.