Circle and a line

the circle centered on the x -axis tangentially touching the line $r \equiv x + y = 0$ ,passing through the point $(2, - 2)$ has equation $(x - 4)^2 + y^2 = 8$ . Note that the line $y + 2 = x - 2$ is perpendicular(normal) to $r$ at $(2,-2)$ touching the x-axis at $(4,0)$ , and the distance between $(2,-2)$ and $(4,0)$ is $\sqrt{8}$ . This give us the equation of the circle,and having this equation is clear that the greatest integral value for $\alpha$ is $\boxed{6}$ .

Since y=- x is the tangent at (2,-2), the radius from that point will be on the line y= - 1/(-1)(x-2) - 2=x-4.
But the center is on x-axis, y=0, so (4,0) is the center, $r^2=(-2+4)^2+2^2= 8.$
The point on the circle whose center is on the x-axis furthest from (0,0), is where circle intersects x-axes, furthest from (0,0).
$=4+r=4+\sqrt8=6.8284,\ so\ \alpha=\Large \ \ \ \color{#D61F06}{6}.$