Let A and B be two distinct points on the parabola $y^{2} = 4x.$ If the axis of the parabola touches a circle of radius r having AB as its diameter, then the slope of the line joining A and B can be $(1) -1/r (2) 1/r (3) 2/r (4) -2/r$

$y^2 = 4x$ $A, B \in \text{parabola}, \quad A \neq B$ $AB \text{ is a diameter}, \quad \text{radius} = r$ $\text{Axis of } y^2 = 4x \text{ is the } x\text{-axis } (y = 0)$ $\text{Axis touches the circle} \Rightarrow \text{tangent}$ $\text{Find the slope of } AB$ $y^2 = 4x \ \Rightarrow \ a = 1$ $\text{vertex } (0,0), \quad \text{focus } (1,0), \quad \text{axis } y = 0$ $A = (t_1^2, 2t_1), \quad B = (t_2^2, 2t_2)$ $t_1 \neq t_2$ $M = \left(\frac{t_1^2+t_2^2}{2},\ t_1+t_2\right)$ $r = \frac{1}{2}|AB|$ $|AB|^2 = (t_1^2-t_2^2)^2 + 4(t_1-t_2)^2$ $= (t_1-t_2)^2\left[(t_1+t_2)^2 + 4\right]$ $r = \frac{1}{2}|t_1-t_2|\sqrt{(t_1+t_2)^2+4}$ $\text{Axis } y = 0; \quad d(M, \text{axis}) = |t_1+t_2|$ $\text{Tangency} \Rightarrow |t_1+t_2| = r$ $\Rightarrow |t_1+t_2| = \frac{1}{2}|t_1-t_2|\sqrt{(t_1+t_2)^2+4}$ $(t_1+t_2)^2 = \frac{1}{4}(t_1-t_2)^2\left[(t_1+t_2)^2+4\right]$ $4S^2 = D^2(S^2+4)$ $m = \frac{2t_1-2t_2}{t_1^2-t_2^2} = \frac{2}{t_1+t_2} = \frac{2}{S}$ $r = |t_1+t_2| = |S| \ \Rightarrow \ S = \pm r$ $m = \frac{2}{S} = \frac{2}{\pm r}$ $m = \pm \frac{2}{r}$ $D^2 = \frac{4S^2}{S^2+4} > 0$ $\text{so } t_1, t_2 \text{ are real and distinct for every } S \neq 0$ $m = \frac{2}{r} \ (3), \qquad m = -\frac{2}{r} \ (4)$ $|m| = \frac{2}{r}$