We have $A=(0,0)$ and $B=(4,-4)$. The parametric equation of the line between those two points is given by the points $$(x,y)=(0,0)+t(4,-4)=t(4,-4),$$
for $t\in[0,1]$. When $t=0$ we get point $A$, when $t=1$ we get point $B$.
Because $C$ is $3$ units from $A$ along the line of length $\sqrt{32}$ from $A$ to $B$, the appropriate parameter value to get point $C$ is $$t_C=\frac{3}{\sqrt{32}}.$$ Similarly, since $D$ is $5$ units from $B$ ($\sqrt{32}-5$ units from $A$) along the line, the appropriate parameter value to get point $D$ is $$t_D=\frac{\sqrt{32}-5}{\sqrt{32}}.$$
The midpoint (call it $E$) is found by taking the average of these two $t$:
$$E=\frac{\sqrt{32}-2}{2\sqrt{32}}(4,-4)\approx (1.29,-1.29)$$
Don't know that it's much simpler than calculating the pairwise intersections, then the distances to the third center, but the following gives a symmetric condition using complex numbers.
Let $\,a,b,c\,$ be the complex numbers associated with points $A,B,C$ in a complex plane centered at the centroid of $ABC\,$, so that $a+b+c=0\,$.
The point of intersection $z$ of the three circles (if it exists) must satisfy the $3$ equations similar to:
$$
|z-a|^2=R_A^2 \;\;\iff\;\;(z-a)(\bar z - \bar a) = R_A^2 \;\;\iff\;\;|z|^2 - z \bar a - \bar z a + |a|^2 = R_A^2 \tag{1}
$$
Writing $(1)$ for $a,b,c$ and summing the $3$ equations up:
$$
\require{cancel}
3\,|z|^2 - \cancel{z \sum_{cyc} \bar a} - \bcancel{\bar z \sum_{cyc} a} + \sum_{cyc}|a|^2 = \sum_{cyc} R_A^2 \;\;\implies\;\; |z|^2 = \frac{1}{3}\left(\sum_{cyc} R_A^2-\sum_{cyc}|a|^2\right) =R^2 \tag{2}
$$
Substituting $(2)$ back into each of $(1)\,$:
$$
-|z|^2 + z \bar a + \bar z a - |a|^2 = - R_A^2 \;\;\iff\;\; z \cdot \bar a + \bar z \cdot a = |a|^2+R^2-R_A^2 \tag{3}
$$
Considering $(3)$ as a system of linear equations in $z, \bar z\,$, the condition for it to have solutions is:
$$
\left|
\begin{matrix}
\;\bar a \;&\; a \;&\; |a|^2+R^2-R_A^2\; \\
\;\bar b \;&\; b \;&\; |b|^2+R^2-R_B^2\; \\
\;\bar c \;&\; c \;&\; |c|^2+R^2-R_C^2\;
\end{matrix}
\right| \;\;=\;\; 0
$$
Best Answer
An absolutely formal proof that the circles cover the unit square would proceed by partitioning the unit square into convex polygons, then showing that each polygon has all its vertices within one of the circles (not necessarily on the boundary). Since the polygons are convex, their vertices lying within a circle implies that their whole body (edges and interior) lies within the same circle too.
And you can in fact do a little better by replacing $\frac{\sqrt5}8$ with $\frac1{1+\sqrt7}$ and using this covering:
If the square is origin-centred, the circle centres are at $(0,0),\left(\pm\frac12,0\right)$ and $\left(\pm\frac14,\pm\frac{2+\sqrt7}{12}\right)$. The proof using the convex partition method is left as an exercise.