7. Let $a$ be a real number, and consider the equation in $z$
$$
\left(z^{2}-2 z+5\right)\left(z^{2}+2 a z+1\right)=0
$$

It has four distinct roots, and the four points corresponding to these roots in the complex plane are concyclic. Then the range of the real number $a$ is
7. $\{-3\} \cup(-1,1)$.

From $z^{2}-2 z+5=0 \Rightarrow z_{1,2}=1 \pm 2 \mathrm{i}$;
From $z^{2}+2 a z+1=0 \Rightarrow(z+a)^{2}=a^{2}-1$.
According to the problem, this equation has four distinct roots, so
$$
a^{2}-1 \neq 0 \Rightarrow a \neq \pm 1 \text {. }
$$

We discuss the cases separately.
(1) If $a^{2}-1>0$, i.e., $a>1$ or $a<-1$, then
$$
z_{3,4}=-a \pm \sqrt{a^{2}-1} \text {. }
$$

To ensure that these four points are concyclic, according to the power of a point theorem, we need
$$
\left|1-z_{1}\right|\left|1-z_{2}\right|=\left|1-z_{3}\right|\left|1-z_{4}\right|,
$$

i.e.,
$$
\begin{aligned}
4 & =\left|1-\left(-a+\sqrt{a^{2}-1}\right)\right|\left|1-\left(-a-\sqrt{a^{2}-1}\right)\right| \\
& =\left|(1+a)^{2}-\left(a^{2}-1\right)\right| .
\end{aligned}
$$

Solving this, we get $a=-3(a=1$ is discarded $)$.
(2) If $a^{2}-1<0$, i.e., $-1<a<1$, then $z_{3,4}=-a \pm \sqrt{1-a^{2}} \mathrm{i}$.
Thus, $z_{1}, z_{2}, z_{3}, z_{4}$ form an isosceles trapezoid, which means the four points are concyclic.
In summary, the real number $a \in\{-3\} \cup(-1,1)$.