1. Set up the circle

Since the circle touches both coordinate axes, its center must be at equal distance from both axes. So let the center be $C(r,r)$ with radius $r$.

Therefore, the points of contact with the axes are: $A(r,0), \quad B(0,r)$ So the line $AB$ passes through $(r,0)$ and $(0,r)$.

Its equation is $x+y=r$

2. Use the condition that $P(\alpha,\beta)$ lies on the circle

The circle equation is $(x-r)^2+(y-r)^2=r^2$ Since $P(\alpha,\beta)$ lies on it, $(\alpha-r)^2+(\beta-r)^2=r^2$ Expanding, $\alpha^2-2\alpha r+r^2+\beta^2-2\beta r+r^2=r^2$ $\alpha^2+\beta^2-2r(\alpha+\beta)+r^2=0 \quad ...(1)$

3. Use the distance from $P$ to line $AB$

Given that $Q$ is the foot of perpendicular from $P$ to $AB$, we have $PQ = \text{distance from }(\alpha,\beta)\text{ to }x+y-r=0$ So, $PQ=\frac{|\alpha+\beta-r|}{\sqrt{2}}$ Given $PQ=11$ and $P$ is above the line $AB$, hence $\alpha+\beta-r>0$ Therefore, $\frac{\alpha+\beta-r}{\sqrt{2}}=11$ $\alpha+\beta-r=11\sqrt{2}$ $r=\alpha+\beta-11\sqrt{2} \quad ...(2)$

4. Substitute into the circle condition

From (1): $\alpha^2+\beta^2-2r(\alpha+\beta)+r^2=0$ Let $s=\alpha+\beta$ Then $\alpha^2+\beta^2=s^2-2\alpha\beta$ So (1) becomes $s^2-2\alpha\beta-2rs+r^2=0$ $2\alpha\beta=s^2-2rs+r^2=(s-r)^2$ Hence, $2\alpha\beta=(\alpha+\beta-r)^2$ But from the distance condition, $\alpha+\beta-r=11\sqrt{2}$ So, $2\alpha\beta=(11\sqrt{2})^2=121\cdot 2$ Thus, $\alpha\beta=121$

5. Final answer

$\boxed{121}$