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(D) Let the other end of the diameter be $B(\alpha, \beta)$.The circle has $AB$ as its diameter,so its equation is $(x - p)(x - \alpha) + (y - q)(y -

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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 circle has $AB$ as its diameter,so its equation is $(x - p)(x - \alpha) + (y - q)(y - \beta) = 0$.Expanding this,we get $x^2 - (p + \alpha)x + p\alpha + y^2 - (q + \beta)y + q\beta = 0$,or $x^2 + y^2 - (p + \alpha)x - (q + \beta)y + (p\alpha + q\beta) = 0$.Since the circle touches the $x$-axis,the $y$-coordinate of the center is equal to the radius,or the discriminant of the equation when $y=0$ is zero.Setting $y = 0$,we get $x^2 - (p + \alpha)x + (p\alpha + q\beta) = 0$.For the circle to touch the $x$-axis,the discriminant $D = 0$.$D = (p + \alpha)^2 - 4(p\alpha + q\beta) = 0$.$p^2 + 2p\alpha + \alpha^2 - 4p\alpha - 4q\beta = 0$.$p^2 - 2p\alpha + \alpha^2 - 4q\beta = 0$.$(p - \alpha)^2 = 4q\beta$.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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