I found out that we can define a circle equation as follows: $$\begin{cases}x(t)=\dfrac{t}{t^2+(kt+b)^2},\\y(t)=\dfrac{kt+b}{t^2+(kt+b)^2},\end{cases}$$ where $k, b$ are real numbers. For example, if we set $b=1, k=-1$, the parametric equation seems to correspond — except for the $t$ such that denominator is zero — to a circle $$\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=\frac{1}{2}$$.
I want to understand why is this parametrization valid, and how can we get circle center and radius using the corresponding line equation $y=kx+b$.
In fact, I've been doing a textbook exercise involving complex numbers, and the first part of the exercise asked to prove that $f(z)=\overline{z}^{-1}$ is a self-inverse, that is $f(f(z))=z$, and the second part asked what the curve of the image of this function with the domain restricted to $z=t+(k\cdot t+b)i$, for some fixed constants $k,b$ and a parameter $t$, was. Such complex numbers $z$ can be visualized with the line equation $y=k\cdot x + b$. The parametric equation of the circle corresponds to a real and imaginary parts of an image of $f$. Thus, as it was a part of an exercise, I am especially interested in explaining this relation between a circle and a line using the property that $f$ is an involution. Other explanations are welcome, too.
Squaring parametric equations and adding them gives $$x^2+y^2=\frac {t^2+(kt+b)^2}{(t^2+(kt+b)^2)^2}=\frac 1{t^2+(kt+b)^2}.$$ Hence parametric equations can be rewritten as $$x=t{(x^2+y^2)}, \quad y=(kt+b)(x^2+y^2).$$ Eliminating $t$ and rearranging gives $$x^2+y^2+\frac kb x - \frac1b y=0\\\left(x+\frac k{2b}\right)^2+\left(y-\frac 1{2b}\right)^2=\left(\frac{\sqrt{k^2+1}}{2b}\right)^2$$ So centre is $$\left(-\frac k{2b}, \frac 1{2b}\right)$$ and radius is $$\frac {\sqrt{k^2+1}}{2b}.$$