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(D) Let the other end of the diameter be $B(\alpha, \beta)$.The center of the circle is the midpoint of $AB$,which is $(\frac{p+\alpha}{2}, \frac{q+\b

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$(x-p)^2=4qy$

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(D) Let the other end of the diameter be $B(\alpha, \beta)$.The center of the circle is the midpoint of $AB$,which is $(\frac{p+\alpha}{2}, \frac{q+\beta}{2})$.The radius of the circle is the distance from the center to the $x$-axis,which is $|\frac{q+\beta}{2}|$.Since the circle passes through $A(p, q)$,the distance from the center to $A$ must equal the radius:$(\frac{p+\alpha}{2} - p)^2 + (\frac{q+\beta}{2} - q)^2 = (\frac{q+\beta}{2})^2$$(\frac{\alpha-p}{2})^2 + (\frac{\beta-q}{2})^2 = (\frac{q+\beta}{2})^2$$(\alpha-p)^2 + (\beta-q)^2 = (q+\beta)^2$$(\alpha-p)^2 + \beta^2 - 2\beta q + q^2 = q^2 + 2\beta q + \beta^2$$(\alpha-p)^2 = 4\beta q$Replacing $(\alpha, \beta)$ with $(x, y)$,the locus is $(x-p)^2 = 4qy$.

$A$ variable circle passes through the fixed point $A(p, q)$ and touches the $x$-axis. The locus of the other end of the diameter through $A$ is

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