An elementary way to deduce this equation might be to observe that your circle has equation
$$
r^2 = (x - a)^2 + (y - b)^2 = \color{red}{(x-a)}\color{green}{(x-a)} + \color{red}{(y - b)}\color{green}{(y - b)}
$$
Since you want a line with exactly one point $(c,d)$ on the circle, it is reasonable to just evaluate the green part at $(c,d)$.
The more advanced answer is: actually, this is a direct application of the general formula for the tangent line of an algebraic curve.
For the purpose of this answer, let us define an algebraic curve as the zero locus (set) in $\Bbb{P^2(R)}$ of a homogeneous polynomial $p(X,Y,Z) \in \Bbb{R}[X,Y,Z]$ -- note that we can always get a curve in the affine plane by setting $Z = 1$, i.e. considering $p(X,Y,1)$. Then by Euler's theorem we know that
$$
\frac{\partial p}{\partial X}(X,Y,Z) \cdot X +
\frac{\partial p}{\partial Y}(X,Y,Z) \cdot Y +
\frac{\partial p}{\partial Z}(X,Y,Z) \cdot Z =
n p(X,Y,Z)
$$
where $n$ is the degree of $p$. Thus we obtain the homogeneous equation of the line tangent to the curve $p(X,Y,Z) = 0$ at a point $(r,s,t)$ as
$$
\frac{\partial p}{\partial X}(r,s,t) \cdot X +
\frac{\partial p}{\partial Y}(r,s,t) \cdot Y +
\frac{\partial p}{\partial Z}(r,s,t) \cdot Z =
n p(r,s,t) = 0 \tag{1} \label{eq:tangent}
$$
Now consider the polynomial $f(X,Y,Z) = (X - aZ)^2 + (Y - bZ)^2 - r^2 Z^2$, so that $f(X,Y,Z) = 0$ is the homogeneous equation of your circle. Then
$$
\begin{align}
\frac{\partial f}{\partial X} &= 2(X - aZ) \\
\frac{\partial f}{\partial Y} &= 2(Y - bZ) \\
\frac{\partial f}{\partial Z} &= -2a(X - aZ) - 2b(Y - bZ) - 2r^2 Z \\
\end{align}
$$
Then we can use \eqref{eq:tangent} to find the equation of the affine tangent line at $(c,d)$:
$$
\begin{align}
0 &=
\frac{\partial f}{\partial X}(c,d,1) \cdot x +
\frac{\partial f}{\partial Y}(c,d,1) \cdot y +
\frac{\partial f}{\partial Z}(c,d,1) \cdot z \\
&= 2(c - a) x + 2(d - b) y - 2a(c - a) - 2b(d - b) - 2r^2 \\
&= 2\big( (c - a)(x - a) + (d - b)(y - b) - r^2 \big)
\end{align}
$$
