1. Circle and endpoints of a diameter

The circle is $x^2+y^2=9,$ so its center is $O(0,0)$ and radius is $3$.

Let the endpoints of a diameter be $P(x_1,y_1), \quad Q(x_2,y_2).$ Since $PQ$ is a diameter of a circle centered at the origin, the midpoint is the origin, hence $Q=(-x_1,-y_1).$ So we may write $P(x,y), \quad Q(-x,-y),$ with $x^2+y^2=9.$

2. Perpendicular distance from a point to the line

The given line is $x+y=2 \quad \text{or} \quad x+y-2=0.$

Distance of point $(x_0,y_0)$ from the line $x+y-2=0$ is $\frac{|x_0+y_0-2|}{\sqrt{1^2+1^2}}=\frac{|x_0+y_0-2|}{\sqrt{2}}.$

Therefore, $\alpha=\frac{|x+y-2|}{\sqrt{2}},$ $\beta=\frac{|-x-y-2|}{\sqrt{2}}=\frac{|x+y+2|}{\sqrt{2}}.$

Thus,

3. Introduce a variable

Let $s=x+y.$ Then \alpha\beta=\frac{|s-2|\,|s+2|}{2}=rac{|s^2-4|}{2}.

Now we need the possible range of $s=x+y$ given that $x^2+y^2=9$.

By Cauchy-Schwarz, $(x+y)^2 \le (1^2+1^2)(x^2+y^2)=2\cdot 9=18.$ So $-3\sqrt{2} \le s \le 3\sqrt{2}.$ Hence $s^2 \le 18.$

4. Maximize the product

We must maximize $\frac{|s^2-4|}{2}$ for $0\le s^2\le 18$.

Let $t=s^2$. Then $0\le t\le 18$, and $\alpha\beta=\frac{|t-4|}{2}.$ This is maximized when $t$ is farthest from $4$ in the interval $[0,18]$, i.e. at $t=18.$ So, $\max(\alpha\beta)=\frac{|18-4|}{2}=\frac{14}{2}=7.$

5. Check attainability

$t=18$ means $(x+y)^2=18,$ which is possible when $x=y=\pm \frac{3}{\sqrt{2}}$. Thus the maximum is attained.

Therefore, the required maximum value is $\boxed{7}.$