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(D) The focus of the parabola $y^2 = 2px$ is $S = (p/2, 0)$.The directrix of the parabola is $x = -p/2$,or $x + p/2 = 0$.Since the circle is centered

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$(A)$ and $(B)$ both

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(D) The focus of the parabola $y^2 = 2px$ is $S = (p/2, 0)$.The directrix of the parabola is $x = -p/2$,or $x + p/2 = 0$.Since the circle is centered at $(p/2, 0)$ and touches the directrix,its radius $r$ is the distance from $(p/2, 0)$ to $x = -p/2$,which is $r = |p/2 - (-p/2)| = |p| = p$.The equation of the circle is $(x - p/2)^2 + y^2 = p^2$.Substitute $y^2 = 2px$ from the parabola into the circle equation:$(x - p/2)^2 + 2px = p^2$$x^2 - px + p^2/4 + 2px = p^2$$x^2 + px - 3p^2/4 = 0$Using the quadratic formula,$x = \frac{-p \pm \sqrt{p^2 - 4(1)(-3p^2/4)}}{2} = \frac{-p \pm \sqrt{4p^2}}{2} = \frac{-p \pm 2p}{2}$.So,$x = p/2$ or $x = -3p/2$.For $x = p/2$,$y^2 = 2p(p/2) = p^2$,so $y = \pm p$.Thus,the points of intersection are $(p/2, p)$ and $(p/2, -p)$.

Consider a circle with its centre lying on the focus of the parabola $y^2 = 2px$ such that it touches the directrix of the parabola. Then,a point of intersection of the circle and the parabola is

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