The truth is that I can't create problems. (Corrected once)

Given that

$\begin{aligned} x^2 + y^2 - 2x - 4y + 4 & = 0 \\ x^2 - 2x + 1 + y^2 - 4y + 4 - 1 & = 0 \\ (x-1)^2 + (y-2)^2 & = 1 & \small \color{#3D99F6} \text{We can rewrite as} \\ \cos^2 \theta + \sin^2 \theta & = 1 & \small \color{#3D99F6} \text{where }x-1 = \cos \theta, \ y-2 = \sin \theta \end{aligned}$

Now consider:

$\begin{aligned} f(x,y) & = x^2 - y^2 + 2\sqrt 3 xy - (4\sqrt 3+2)x + (4-2\sqrt 3)y - (4\sqrt 3+3) \\ & = x^2 - 2x + 1 - (y^2 - 4y + 4) + 2\sqrt 3 (xy - 2x - y) - 4\sqrt 3 \\ & = (x-1)^2 - (y-2)^2 + 2\sqrt 3(x-1)(y-2) - 8\sqrt 3 & \small \color{#3D99F6} \text{Put }x-1 = \cos \theta, \ y-2 = \sin \theta \\ & = \cos^2 \theta - \sin^2 \theta + 2\sqrt 3\cos \theta \sin \theta - 8 \sqrt 3 \\ & = \cos (2\theta) + \sqrt 3 \sin (2\theta) - 8 \sqrt 3 \\ & = 2\left(\frac 12 \cos(2\theta) + \frac {\sqrt 3}2\sin (2\theta)\right) - 8\sqrt 3 \\ & = 2 \sin \left(2\theta + \frac \pi 6\right) - 8\sqrt 3 \end{aligned}$

We note that $|f(x,y)|$ is maximum, when $f(x,y)$ is minimum or having the largest negative value. And that is when $\sin \left(2\theta + \frac \pi 6\right) = -1$ . Therefore, $\max |f(x,y)| = \left|-2-8\sqrt 3\right| = 8\sqrt 3 + 2$ . $\implies ab^2 + c = 8(3^2) + 2 = \boxed{74}$ .