What is the name/family behind the following high-pass IIR filter:
$ T = 1/f_s \\ f_c = 5 \\ c_1 = \displaystyle \frac{1}{1 + \tan(\pi f_cT)}\\ c_2 = \displaystyle \frac{1 - \tan(\pi f_cT)}{1 + \tan(\pi f_cT)} \\ b = \left[c_1, -c_1\right] \\ a = \left[1, -c_2\right] $
Where $f_s$ is the sampling rate and $f_c$ the cut-off frequency.
This is a first-order discrete-time Butterworth filter, which can be obtained via the bilinear transform from the following analog prototype filter (with normalized cut-off frequency):
$$H(s)=\frac{s}{1+s}\tag{1}$$
The bilinear transform replaces $s$ by $k\frac{z-1}{z+1}$, where $k$ is chosen such that the discrete-time filter has the specified cut-off frequency $f_c$, resulting in
$$k=\frac{1}{\tan(\pi f_c T)}\tag{2}$$
where $T=1/f_s$ is the sampling interval.
So if you replace $s$ in $(1)$ by
$$s=\frac{1}{\tan(\pi f_c T)}\frac{z-1}{z+1}\tag{3}$$
you obtain the corresponding discrete-time filter with the coefficients given in your question.
EDIT:
OK, here is how you obtain your final coefficients. Let $a=\tan(\pi f_c T)$. Plugging $(3)$ into $(1)$ then gives for the transfer function of the discrete-time filter
$$\begin{align}G(z)=\frac{\frac{1}{a}\frac{z-1}{z+1}}{1+\frac{1}{a}\frac{z-1}{z+1}}&=\frac{z-1}{a(z+1)+z-1}\\&=\frac{z-1}{(1+a)z-(1-a)}\\&=\frac{1}{1+a}\frac{z-1}{z-\frac{1-a}{1+a}}\end{align}\tag{4}$$
which is exactly in the given form with $c_1=1/(1+a)$ and $c_2=(1-a)/(1+a)$.
