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

$\begin{array}{l}z=\frac{1}{a+bit},t\in R,t\ne 0\\ z=\frac{a-bit}{a^2+b^2{t}^2}=x+iy \\ \frac{x}{y}=\frac{-a}{bt}\Rightarrow t=\frac{-ay}{bx} \\ \sin ce x=\frac{a}{a^2+b^2{t}^2}\\ \Rightarrow x=a\left(x^2+y^2\right)\\ x^2+y^2=\frac{x}{a}\end{array}$

$\text{ Circle with radius }=\frac{1}{2a},\text{ centre }=\left(\frac{1}{2a},0\right)$

$\text{ Also if }b=0,a\ne 0\Rightarrow z=\frac{1}{a}\text{ purely real lies on }\mathrm{x}\text{ -axis }$

$\text{ if }b=0,a\ne 0\Rightarrow z=\frac{1}{bit}\text{ purely imaginary lies on }\mathrm{y}\text{ -axis }$