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(C) Let the center of the circle be $O(\alpha, \beta)$ and the other end of the diameter through $A(1,1)$ be $Q(h, k)$.Since the circle touches the $X

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$(x-1)^2=4y$

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(C) Let the center of the circle be $O(\alpha, \beta)$ and the other end of the diameter through $A(1,1)$ be $Q(h, k)$.Since the circle touches the $X$-axis,the radius is $|\beta|$. Thus,the equation of the circle is $(x-\alpha)^2 + (y-\beta)^2 = \beta^2$.Since $A(1,1)$ lies on the circle,we have $(1-\alpha)^2 + (1-\beta)^2 = \beta^2$,which simplifies to $(1-\alpha)^2 + 1 - 2\beta = 0$,so $2\beta = (1-\alpha)^2 + 1$.Since $O(\alpha, \beta)$ is the midpoint of the diameter $AQ$,we have $\alpha = \frac{h+1}{2}$ and $\beta = \frac{k+1}{2}$.Substituting these into the equation $2\beta = (1-\alpha)^2 + 1$:$k+1 = (1 - \frac{h+1}{2})^2 + 1$$k+1 = (\frac{2-h-1}{2})^2 + 1$$k+1 = \frac{(1-h)^2}{4} + 1$$k = \frac{(h-1)^2}{4}$$4k = (h-1)^2$.Replacing $(h, k)$ with $(x, y)$,the locus is $(x-1)^2 = 4y$.

If a circle passing through $A(1,1)$ touches the $X$-axis,then the locus of the other end of the diameter through $A$ is

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