Coordinate Coordinate!!!

Let the center of the circle be $O$ and its radius $r$ . From the coordinates of $A$ , $B$ , $C$ and $D$ , we know that $O$ has a positive $x$ -coordinate $a$ and negative $y$ -coordinate $b$ , that is $O(a,-b)$ . And we note that $a=\dfrac{x+x+1}{2} = x + \frac{1}{2}$ .

From the coordinates, we have:

$\begin{cases} OA: & a^2+b^2 & = r^2 \\ OB: & (a+1)^2+(b-1)^2 & = r^2 \\ OC: & (y+b)^2 +\frac{1}{4} & = r^2 \end{cases}$

\(\begin{array} {} OA = OB: & a^2+b^2 = (a+1)^2+(b-1)^2 = a^2 + 2a + 1 + b^2 -2b + 1 \\ & \Rightarrow 2a-2b+2 = 0 \quad \Rightarrow b = a + 1 = x + \frac{3}{2} \\ OA= OC: & a^2+b^2 = (y+b)^2 +\frac{1}{4} = y^2 + 2by + b^2 + \frac{1}{4} \\ & \Rightarrow y^2 + 2by + \frac{1}{4} - a^2 = 0 \\ & \quad y^2 + 2by + \frac{1}{4} - \left(x^2 + x + \frac{1}{4} \right) = 0 \\ & \quad y^2 + (2x+3) y - x^2 - x = 0 \end{array} \)

For integer value of $y \Longrightarrow (2x+3)^2 + 4(x^2+x) = 8x^2 + 16x + 9$ must be a perfect square. And the two smallest $x > 0$ , when this occurs, are:

$\begin{cases} x = 5 & \Rightarrow a = 5.5 & \Rightarrow b = 6.5 & \Rightarrow r_1^2 = 5.5^2 + 6.5^2 & = 72.5 \\ x = 34 & \Rightarrow a = 34.5 & \Rightarrow b = 35.5 & \Rightarrow r_2^2 = 34.5^2+35.5^2 & = 2450.5 \end{cases}$

$\Rightarrow r_1^2 + r_2^2 = 72.5 + 2450.5 = \boxed{2523}$