Greatest value of $\alpha$

Since the line $x+y = 0$ is a tangent, the line normal to the tangent (this line will pass through the center of the circle) will have equation of $y = x+C$

to determine $C$ , notices that (2,-2) is a point of that normal. Thus

$-2 = 2+C$

$C = -4$

The centre of the circle is on the X-axis, so to figure out the coordinate of the centre:

$y = x -4$

$0 = x -4$

$x = 4$

Thus the centre of the circle is located at $(4,0)$ .

To complete the equation of the circle, notice that (2,-2) is a point on the circle.

$\frac{(x-4)^2}{k^2}+\frac{y^2}{k^2} = 1$

$\frac{(-2)^2}{k^2}+\frac{(-2)^2}{k^2} = 1$

$\frac{4}{k^2}+\frac{4}{k^2} = 1$

$k^2 = 8$

$k = 2\sqrt{2}$

As $(\alpha,\beta)$ is on the circle, maximum value of $\alpha$ occurs at $(\alpha,0)$ which gives

$\frac{(\alpha-4)^2}{(2\sqrt{2})^2}+\frac{(0)^2}{(2\sqrt{2})^2} = 1$

$\frac{(\alpha-4)^2}{(2\sqrt{2})^2} = 1$

$\frac{(\alpha-4)^2} = {(2\sqrt{2})^2}$

$\alpha = 4 \pm 2\sqrt{2}$

These points are where the circle intersects with the x-axis. We take the larger value, $\alpha = 4+ 2\sqrt{2}$ in this case for the maximum.