The side $$𝐴𝐵$$ of a square $$𝐴𝐵𝐶𝐷$$ is $$1$$ and it is also a chord of a circle $$𝑆$$. The side $$𝐶𝐷$$ does not intersect $$𝑆$$. The length of the tangent $$𝐶𝐾$$, drawn from $$𝐶$$ to $$𝑆$$ at the point $$𝐾$$ is $$2$$. If $$𝑑$$ is the diameter of $$𝑆$$, then calculate $$d^{2}$$.

Let us place square $$ABCD$$ on a coordinate plane so that
$$A(0,0),\; B(1,0),\; C(1,1),\; D(0,1).$$
Thus $$AB=1$$ lies on the $$x$$-axis.

Because $$AB$$ is a chord of the required circle $$S$$, its centre $$O(x_0,y_0)$$ must lie on the perpendicular bisector of $$AB$$, i.e. on the vertical line $$x=\tfrac12$$.
Write $$O\equiv\left(\tfrac12,\;h\right)$$ where $$h$$ is to be found.

1. Radius in terms of $$h$$:
$$r^2 = OA^2 = \left(\tfrac12\right)^2 + h^2 = 0.25 + h^2.$$ $$-(1)$$

2. Power of point $$C(1,1)$$ with respect to $$S$$.
The tangent length from $$C$$ is given as $$CK=2$$, therefore
$$\text{Power}(C) = CK^2 = 4.$$

Using the power-of-a-point theorem
$$\text{Power}(C)=OC^2-r^2.$$
Compute $$OC^2$$:
$$OC^2=\left(1-\tfrac12\right)^2+(1-h)^2 = 0.25 + (1-h)^2.$$

Hence
$$\left[0.25 + (1-h)^2\right] - \left[0.25 + h^2\right] = 4.$$
The $$0.25$$ terms cancel, giving
$$(1-h)^2 - h^2 = 4.$$ Expand and simplify:
$$1 - 2h + h^2 - h^2 = 4 \;\Longrightarrow\; 1 - 2h = 4 \;\Longrightarrow\; h = -\dfrac32.$$ So the centre lies $$1.5$$ units below the side $$AB$$.

3. Radius and diameter.
From (1):
$$r^2 = 0.25 + \left(-\dfrac32\right)^2 = 0.25 + 2.25 = 2.5 = \dfrac52.$$ Therefore
$$r = \sqrt{\dfrac52},\qquad d = 2r = 2\sqrt{\dfrac52}= \sqrt{10}.$$

4. Verification of the given geometric conditions.
The highest point of the circle is at $$y = h + r = -1.5 + 1.581\ldots \approx 0.08$$, which is below the line $$CD$$ ( $$y=1$$ ). Hence $$CD$$ does not meet the circle, exactly as stated in the problem.

Finally
$$d^{2} = (\sqrt{10})^{2} = 10.$$

Answer: 10