The given circle equation is \(x^2 + y^2 - x - y - 6 = 0\). To find the center and radius of the circle, we need to rewrite it in the standard form.
Start by completing the square:
\(x^2 - x + y^2 - y = 6\)
\(\left(x - \frac{1}{2}\right)^2 - \frac{1}{4} + \left(y - \frac{1}{2}\right)^2 - \frac{1}{4} = 6\)
\(\left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = 6 + \frac{1}{2}\)
\(\left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{13}{2}\)
This represents a circle with center \(\left(\frac{1}{2}, \frac{1}{2}\right)\) and radius \(\sqrt{\frac{13}{2}}\).
The chord equation is \(x + y - 2 = 0\). The equation of the circle's center \(\left(\frac{1}{2}, \frac{1}{2}\right)\) with respect to the chord is:
\(\frac{1}{2} + \frac{1}{2} - 2 = 1 - 2 = -1\)
The perpendicular distance from the center to the line \(x + y = 2\) is:
\(\frac{|1 - 2|}{\sqrt{1^2 + 1^2}} = \frac{1}{\sqrt{2}}\)
The point \((\alpha - 1, \alpha + 1)\) lies on either side of the line.
Substitute \((\alpha - 1, \alpha + 1)\) into the line equation:
\((\alpha - 1) + (\alpha + 1) - 2 = 2\alpha - 2\)
We compare it with the center sign to check which side it lies:
\(2\alpha - 2 = 0 \rightarrow \alpha = 1\). This value of \(\alpha\) makes the point lie on the line.
Thus, \(2\alpha - 2\) should be less than 0 for the point to lie in the larger segment (same side as the center):
\(2\alpha - 2 < 0 \Rightarrow 2\alpha < 2 \Rightarrow \alpha < 1\)
For \(\alpha = -1\), \(2\alpha - 2 = -4\) which implies the point lies further from the line compared to the center's negative side.\)
Thus, the range of \(\alpha\) must be \(-1 < \alpha < 1\).
Hence, the correct option is: \(-1 < \alpha < 1\).