An arrangement of circles in which circles intersect only in angles of $\pi/2$ is called an \emph{arrangement of orthogonal circles}. We show that in the case that no two circles are nested, the intersection graph of such an arrangement is planar. The same result holds for arrangement of circles that intersect in an angle of at most $\pi/2$. For the general case we prove that the maximal number of edges in an intersection graph of an arrangement of orthogonal circles lies in between $4n - O\left(\sqrt{n}\right)$ and $\left(4+\frac{5}{11}\right)n$, for $n$ being the number of circles. Based on the lower bound we can also improve the bound for the number of triangles in arrangements of orthogonal circles to $(3 + 5/9)n-O\left(\sqrt{n}\right)$.

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