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

Here,  $x+iy=\frac{1}{a+ibt}\times \frac{a-ibt}{a-ibt}$

$\therefore x+iy=\frac{a-ibt}{a^2+b^2{t}^2}$

Let    $a\ne 0,b\ne 0$

$\begin{array}{l}\therefore x=\frac{a}{a^2+b^2{t}^2}\text{ and }y=\frac{-bt}{a^2+b^2{t}^2}\\ \frac{y}{x}=\frac{-bt}{a}\Rightarrow t=\frac{ay}{bx}\end{array}$

On putting $x=\frac{a}{a^2+b^2{t}^2}$ ,we get

$\begin{array}{l}x\left(a^2+b^2\cdot \frac{a^2{y}^2}{b^2{x}^2}\right)=a\\ \Rightarrow a^2\left(x^2+y^2\right)=ax\end{array}$

or      $x^2+y^2-\frac{x}{a}=0$           ... (i) 

or     ${\left(x-\frac{1}{2a}\right)}^2+y^2=\frac{1}{4a^2}$

$\therefore$    Option (a) is correct 

For     $a\ne 0\text{ and }b=0$

$x+iy=\frac{1}{a}\Rightarrow x=\frac{1}{a},y=0$

$\Rightarrow z\text{ lies on }X\text{-axis. }$

$\therefore$  Option (d) is correct.