jmc

Since $A$ and $B$ lie on the graph of $y^2=4x$ in the first quadrant, we can let $A=(a^2,2a)$ and $B=(b^2,2b),$ where $a$ and $b$ are positive. Then the center of the circle is the midpoint of $\overline{AB},$ or $$\left(\frac{a^2+b^2}{2},a+b\right).$$

Since the circle is tangent to the $x$-axis, the radius of the circle is $r=a+b.$

The slope of line $AB$ is then $$\frac{2a-2b}{a^2-b^2}=\frac{2(a-b)}{(a+b)(a-b)}=\frac{2}{a+b}=\boxed{\frac{2}{r}}.$$