1. Parametrize the points on the parabola

For the parabola $y^2=4x$, a general point can be written as $A(t_1^2,2t_1), \qquad B(t_2^2,2t_2).$ Since $A,B$ are distinct, $t_1\neq t_2$.

2. Find the circle having $AB$ as diameter

The circle with diameter endpoints $A(x_1,y_1)$ and $B(x_2,y_2)$ has:

• center at the midpoint of $AB$,
• radius equal to half the length of $AB$.

So the center is $\left(\frac{t_1^2+t_2^2}{2},\, t_1+t_2\right).$

The length $AB$ is $AB=\sqrt{(t_1^2-t_2^2)^2+(2t_1-2t_2)^2}.$ Factor: $t_1^2-t_2^2=(t_1-t_2)(t_1+t_2),$ so $AB=|t_1-t_2|\sqrt{(t_1+t_2)^2+4}.$ Hence radius of this circle is $r=\frac{|t_1-t_2|}{2}\sqrt{(t_1+t_2)^2+4}.$

3. Use the condition that the axis touches the circle

The axis of the parabola $y^2=4x$ is the $x$-axis, i.e. $y=0$.

For the line $y=0$ to touch the circle, the perpendicular distance from the center to the line must equal the radius.

The center has $y$-coordinate $t_1+t_2$, so distance from center to $y=0$ is $|t_1+t_2|.$ Thus tangency gives $|t_1+t_2|=r.$

Now substitute the expression for $r$: $|t_1+t_2|=\frac{|t_1-t_2|}{2}\sqrt{(t_1+t_2)^2+4}.$

Square both sides: $4(t_1+t_2)^2=(t_1-t_2)^2\big((t_1+t_2)^2+4\big).$

Let $s=t_1+t_2,\qquad d=t_1-t_2.$ Then $4s^2=d^2(s^2+4).$ Also from tangency, since $r=|s|$, $s^2=r^2.$

So $4r^2=d^2(r^2+4),$ which gives $d^2=\frac{4r^2}{r^2+4}.$

4. Find the slope of line joining $A$ and $B$

Slope of chord $AB$ is m=\frac{2t_1-2t_2}{t_1^2-t_2^2}=rac{2(t_1-t_2)}{(t_1-t_2)(t_1+t_2)}=\frac{2}{t_1+t_2}=rac{2}{s}, provided $t_1\neq t_2$, which is true.

Since tangency gave $|s|=r,$ we have $s=\pm r.$ Therefore $m=\frac{2}{\pm r}=\pm \frac{2}{r}.$

So the possible slopes are $\frac{2}{r},\; -\frac{2}{r}.$

5. Check options

• A: $-\frac{1}{r}$ ❌
• B: $\frac{1}{r}$ ❌
• C: $\frac{2}{r}$ ✅
• D: $-\frac{2}{r}$ ✅

Thus the correct options are C and D.