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Two graphs having the same spectrum are said to be cospectral. Two graphs such that the absolute values of their nonzero eigenvalues coincide are singularly cospectral graphs. Cospectrality implies singular cospectrality, but the converse may be false. In this paper, we present sufficient conditions for two circulant graphs, with an even number of vertices, to be noncospectral singularly cospectral graphs. In this analysis, we study when a pair of these graphs have the same or distinct inertia. In addition, we show that two singularly cospectral circulant graphs with an odd prime number of vertices are isomorphic.

A signed graph is one that features two types of edges: positive and negative. Balanced signed graphs are those in which all cycles contain an even number of positive edges. In the adjacency matrix of a signed graph, entries can be $0$, $-1$, or $1$, depending on whether $ij$ represents no edge, a negative edge, or a positive edge, respectively. The index of the adjacency matrix of a signed graph $\dot{G}$ is less or equal to the index of the adjacency matrix of its underlying graph $G$, i.e., $\lambda_1(\dot{G}) \le \lambda_1(G)$. Indeed, if $\dot{G}$ is balanced, then $\lambda_1(\dot{G})=\lambda_1(G)$. This inequality becomes strict when $\dot{G}$ is an unbalanced signed graph. Recently, Brunetti and Stani\'c found the whole list of unbalanced signed graphs on $n$ vertices with maximum (resp. minimum) spectral radius. To our knowledge, there has been little research on this problem when unbalanced signed graphs are confined to specific graph classes. In this article, we demonstrate that there is only one unbalanced signed bipartite graph on $n$ vertices with maximum spectral radius, up to an operation on the signed edges known as switching. Additionally, we investigate unbalanced signed complete bipartite graphs on $n$ vertices with a bounded number of edges and maximum spectral radius, where the negative edges induce a tree.