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(D) Given $z = \frac{1}{a+ibt}$.$x+iy = \frac{a-ibt}{a^2+b^2t^2}$.Equating real and imaginary parts,$x = \frac{a}{a^2+b^2t^2}$ and $y = \frac{-bt}{a^2

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$A, C, D$

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(D) Given $z = \frac{1}{a+ibt}$.$x+iy = \frac{a-ibt}{a^2+b^2t^2}$.Equating real and imaginary parts,$x = \frac{a}{a^2+b^2t^2}$ and $y = \frac{-bt}{a^2+b^2t^2}$.If $a 
eq 0$ and $b 
eq 0$,then $a^2+b^2t^2 = \frac{a}{x}$,so $b^2t^2 = \frac{a}{x} - a^2 = \frac{a(1-ax)}{x}$.Also $y^2 = \frac{b^2t^2}{(a^2+b^2t^2)^2} = \frac{a(1-ax)/x}{(a/x)^2} = \frac{x(1-ax)}{a} = \frac{x}{a} - x^2$.Rearranging gives $x^2 - \frac{x}{a} + y^2 = 0$,which is $(x - \frac{1}{2a})^2 + y^2 = (\frac{1}{2a})^2$. This represents a circle with radius $|\frac{1}{2a}|$ and center $(\frac{1}{2a}, 0)$.If $b=0$,then $z = \frac{1}{a}$,so $y=0$,which is the $x$-axis.If $a=0$,then $z = \frac{1}{ibt} = -i(\frac{1}{bt})$,so $x=0$,which is the $y$-axis.Thus,options $A, C, D$ are correct.

Let $a, b \in \mathbb{R}$ and $a^2+b^2 \neq 0$. Suppose $S = \{z \in \mathbb{C} : z = \frac{1}{a+ibt}, t \in \mathbb{R}, t \neq 0\}$,where $i = \sqrt{-1}$. If $z = x+iy$ and $z \in S$,then $(x, y)$ lies on:

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