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Question

(A) Let the other end of the diameter be $B(\alpha, \beta)$.The equation of the circle with diameter $AB$ is:$(x - p)(x - \alpha) + (y - q)(y - \beta)

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the other end of the diameter be $B(\alpha, \beta)$.The equation of the circle with diameter $AB$ is:$(x - p)(x - \alpha) + (y - q)(y - \beta) = 0$$x^{2} + y^{2} - (p + \alpha)x - (q + \beta)y + p\alpha + q\beta = 0$ ... $(i)$Since the circle touches the $x$-axis,the $y$-coordinate of the center is equal to the radius,or the discriminant of the equation obtained by setting $y = 0$ is zero.Setting $y = 0$ in $(i)$:$x^{2} - (p + \alpha)x + (p\alpha + q\beta) = 0$Since it touches the $x$-axis,the discriminant $D = 0$:$(p + \alpha)^{2} - 4(p\alpha + q\beta) = 0$$p^{2} + 2p\alpha + \alpha^{2} - 4p\alpha - 4q\beta = 0$$p^{2} - 2p\alpha + \alpha^{2} = 4q\beta$$(p - \alpha)^{2} = 4q\beta$Replacing $(\alpha, \beta)$ with $(x, y)$,the locus is $(x - p)^{2} = 4qy$.

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

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