Hidden Pythagoras

Let the center of circle with radius $1$ be $O$ and $OD$ be perpendicular to $AC$ . Let $CD=x$ and $\angle DCO = \theta$ . Then $\angle ACB = 2\theta$ . It is easy to see that $AD = OQ=4$ . The $AC = x+4$ . We have:

$\begin{aligned} \tan \angle DCO & = \tan \theta = \frac {OD}{CD} = \frac 1x \\ \tan \angle ACB & = \tan (2\theta) = \frac {AB}{AC} = \frac 8{x+4} \\ \implies \frac 8{x+4} & = \frac {\frac 2x}{1-\frac 1{x^2}} = \frac {2x}{x^2-1} \\ 4x^2 - 4 & = x^2 + 4x \\ 3x^2 - 4x - 4 & = 0 \\ (3x+2)(x-2) & = 0 \\ \implies x & = 2 & \small \blue{\text{Since }x >0} \end{aligned}$

Therefore, $AC = x + 4 = \boxed 6$ .

Let $O_1$ be the center of the circle with radius $1$ , $O_2$ be the center of the circle with radius $4$ , then drop a perpendicular line from $O_1$ to $AB$ , and such that $O_1 D$ is perpendicular to $AB$ . Then we have $O_1D=4$ . And then the area of $∆ABC$ is $4(?)$ , which then have

$4(?)=\frac{(1)(?)+(8)(4)+\sqrt{?^2+64}(1)}{2}$ $?=6$