Slope of Chord Forming Diameter of Circle Touching Parabola Axis

To find the possible slopes of the line joining points $\text{A}$ and $\text{B}$, we proceed with the following steps:

1. Parametric representation of points on the parabola:
   The equation of the given parabola is:
   $y^2 = 4x$
   Let the coordinates of points $\text{A}$ and $\text{B}$ on the parabola be:
   $\text{A} = (x_1, y_1) \quad \text{and} \quad \text{B} = (x_2, y_2)$
   Since both points lie on the parabola, they satisfy:
   $y_1^2 = 4x_1 \implies x_1 = \frac{y_1^2}{4}$
   $y_2^2 = 4x_2 \implies x_2 = \frac{y_2^2}{4}$

2. Slope of the chord $\text{AB}$:
   The slope $m$ of the line segment joining $\text{A}$ and $\text{B}$ is given by:
   $m = \frac{y_2 - y_1}{x_2 - x_1}$
   Substituting $x_1$ and $x_2$:
   $m = \frac{y_2 - y_1}{\frac{y_2^2}{4} - \frac{y_1^2}{4}} = \frac{4(y_2 - y_1)}{(y_2 - y_1)(y_2 + y_1)}$
   Since $\text{A}$ and $\text{B}$ are distinct points, $y_1 \neq y_2$, which simplifies to:
   $m = \frac{4}{y_1 + y_2}$

3. Condition for the circle touching the axis of the parabola:
   The axis of the parabola $y^2 = 4x$ is the line:
   $y = 0 \quad (\text{the } x\text{-axis})$
   The circle has diameter $\text{AB}$ and radius $r$.
   The center of the circle, $\text{C}$, is the midpoint of the diameter $\text{AB}$:
   $\text{C} = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$
   Since the circle touches the axis of the parabola ($y = 0$), the perpendicular distance from the center $\text{C}$ to the line $y = 0$ must be equal to the radius $r$ of the circle:
   $\left|\frac{y_1 + y_2}{2}\right| = r$
   $\implies y_1 + y_2 = \pm 2r$

4. Calculating the slope $m$:
   Substituting $y_1 + y_2 = \pm 2r$ into the expression for the slope $m$:
   $m = \frac{4}{\pm 2r} = \pm \frac{2}{r}$

Thus, the slope of the line joining $\text{A}$ and $\text{B}$ can be $\frac{2}{r}$ or $-\frac{2}{r}$.

Therefore, the correct options are C and D.