Review for Exam 1 Problem 3.23

Given $a\in\mathbb{R},a\not=0$ we are asked to show that the image of a line $y=a$ under inversion is the circle with center $-\frac{i}{2a}$ and radius $\frac{1}{2a}$. I take this to mean the line in question is a horizontal line in the complex plane $z=x+ia$. Under inversion

$$
\frac{1}{z}=\frac{1}{x+ia}=\frac{1}{x+ia}\cdot\frac{x-ia}{x-ia}=\frac{x-ia}{x^2+a^2}=\frac{x}{x^2+a^2}+i\Bigg(\frac{-a}{x^2+a^2}\Bigg)
$$

I'm unsure how to find the equation of a circle centered at $-\frac{i}{2a}$ with radius $\frac{1}{2a}$ using this information.

I'm not sure how your introduced a $y$ into your formula but, once you deal with that, I guess you could show that your real part squared plus your (imaginary part - 1/2a)^2 is the appropriate constant.