Solution. Two circles intersect at an angle $\alpha$ if the angle formed by the tangents at the intersection point is equal to $\alpha$.

If a given circle passes through the points $(-1,0)$ and $(1,0)$, then its center lies on the $y$-axis. Let's denote this center by $O(0, q)$. Then the equation of an arbitrary circle $k$ passing through the points $(-1,0)$ and $(1,0)$ is

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$$k: \quad x^{2}+(y-q)^{2}=q^{2}+1$$

i.e., $k\left((0, q), \sqrt{q^{2}+1}\right)$. Furthermore, two circles $k_{1}\left(O_{1}, r_{1}\right)$ and $k_{2}\left(O_{2}, r_{2}\right)$ intersect at an angle of $90^{\circ}$ if and only if

$$r_{1}^{2}+r_{2}^{2}={\overline{O_{1} O_{2}}}^{2}$$

Let the center of the circle with radius $r>0$ that intersects an arbitrary circle of the form (1) at an angle of $90^{\circ}$ be at the point $(a, b)$. From (2) we have

$$r^{2}+q^{2}+1=a^{2}+(b-q)^{2} \text {, i.e., } r^{2}+1=a^{2}+b^{2}-2 b q$$

The last equality must hold for every $q$, which is possible only if $b=0$. Now, substituting into the last equation, we get $a= \pm \sqrt{r^{2}+1}$. Finally, the center of the desired circle is $\left(-\sqrt{r^{2}+1}, 0\right)$ or $\left(\sqrt{r^{2}+1}, 0\right)$.