Set up coordinates in the plane; then every circle can be expressed as $p(x, y) = x^2 + y^2 + l(x, y) = 0$, where $l(x, y)$ is a polynomial of degree at most one. Also note that for every point $A = (x_A, y_A)$ in the plane, $p(x_A, y_A) = d^2 - r^2$, where $d$ is the distance from $A$ to the center of the circle, and $r$ is the radius of the circle.

Now, for each $i \in \{1, 2, 3, 4\}$, let $p_i(x, y) = x^2 + y^2 + l_i(x, y) = 0$ denote the equation of the corresponding circle with center $O_i$ and radius $r_i$, and let $d_i$ be the distance from $A_i$ to $O_i$. Then the four points $A_1, A_2, A_3, A_4$ all satisfy the equation
$$\sum_{i=1}^{4} \frac{p_i(x, y)}{d_i^2 - r_i^2} = 1. \qquad (1)$$

However, the four points $A_1, A_2, A_3, A_4$ are neither concyclic nor collinear, so (1) is neither a circle nor a line. Hence the coefficient of $x^2 + y^2$ on the left-hand side of (1) must be zero, that is,
$$\sum_{i=1}^{4} \frac{1}{d_i^2 - r_i^2} = 1.$$