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(C) The equation of the parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$. Let the chord be $y = mx + c$. The points of intersection of the line

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(C) The equation of the parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$. Let the chord be $y = mx + c$. The points of intersection of the line $y = mx + c$ and the parabola $y^2 = 8x$ are obtained by substituting $x = (y - c)/m$ into the parabola equation: $y^2 = 8(y - c)/m \Rightarrow my^2 - 8y + 8c = 0$. Let the points be $P(x_1, y_1)$ and $Q(x_2, y_2)$. Then $y_1 + y_2 = 8/m$ and $y_1y_2 = 8c/m$. The midpoint of the chord is $M = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$. The circle with the chord as diameter has radius $R = 4$. The distance from the center of the circle to the axis of the parabola $(y = 0)$ is equal to the radius $4$,so the $y$-coordinate of the center is $4$ (or $-4$). Thus,$\frac{y_1+y_2}{2} = 4 \Rightarrow y_1 + y_2 = 8$. Since $y_1 + y_2 = 8/m$,we have $8/m = 8$,which implies $m = 1$.

$A$ circle of radius $4$,drawn on a chord of the parabola $y^2 = 8x$ as diameter,touches the axis of the parabola. Then,the slope of the chord is

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