For $0 < k < 6,$ the graphs of $\frac{(x - k)^2}{9} + y^2 = 1$ and $\frac{x^2}{9} + y^2 = 1$ intersect at $A$ and $C,$ and have $x$-intercepts at $B$ and $D$ respectively.  Compute the value of $k$ for which $ABCD$ is a square.

[asy]
unitsize(1 cm);

path ellone = xscale(3)*Circle((0,0),1);
path elltwo = shift((24/5,0))*xscale(3)*Circle((0,0),1);
pair A, B, C, D;

A = intersectionpoints(ellone,elltwo)[0];
C = intersectionpoints(ellone,elltwo)[1];
B = (-3 + 24/5,0);
D = (3,0);

draw(ellone);
draw(elltwo);

draw((-3.5,0)--(9,0));
draw((0,-1.5)--(0,1.5));
draw(A--B--C--D--cycle);

label("$A$", A, N, fontsize(10));
label("$B$", B, NW, fontsize(10));
label("$C$", C, S, fontsize(10));
label("$D$", D, NE, fontsize(10));
[/asy]
We have that $D = (3,0)$ and $B = (k - 3,0).$  Hence, the $x$-coordinates of $A$ and $C$ are $\frac{k}{2}.$  The length of diagonal $BD$ is $6 - k,$ so the $y$-coordinate of $A$ is $\frac{6 - k}{2}.$  Hence,
\[\frac{(k/2)^2}{9} + \left( \frac{6 - k}{2} \right)^2 = 1.\]This simplifies to $5k^2 - 54k + 144 = 0,$ which factors as $(k - 6)(5k - 24) = 0.$  Hence, $k = \boxed{\frac{24}{5}}.$