A pair of circles intersect in points $P$ and $Q$. Let $\overline{AB}$ be a segment passing through $P$ having one endpoint on each circle, as shown. If the circles have radii $6$ and $12$, while $PQ=8$, then determine the largest possible length of $\overline{AB}$.
Right triangle $BPQ$ is inscribed in circle $\odot(BPQ)$, so $\angle{PBQ} = \angle{PQB}$. We also have that
$$BQ = \frac{\overarc{BP} + \overarc{PQ}}{2} = \frac{\overarc{PBQ}}{2},$$where $\overarc{BP}$ denotes the minor arc between $B$ and $P$. Thus, if $\theta = \angle{PBQ}$, then
$$\overarc{BPQ} = 4\theta,$$as shown below.
[asy]
/* Made by MRENTHUSIASM */
size(200);
pair A, B, O1, O2, P, Q;
A = (-30, -10);
B = (0, 12);
O1 = (0, 0);
O2 = (0, 8);
P = (-11.174, 10.408);
Q = (-11.174, -8.408);
dot("$A$", A, W);
dot("$B$", B, dir(75));
clip(anglemark(Q, B, A, 80));
draw(arc(O1, 12, 0, 180));
draw(arc(O2, 6, 0, 180));
draw(A--B);
draw(arc(B, 5, 95, 147), dashed);
draw(P--Q, gray);
[/asy]
Note that $PBQ \sim ABC$. Therefore,
$$\frac{\overline{AB}}{6} = \frac{AB}{12} = \frac{AB+8}{8} = \cos\left(\frac{2\theta}{\pi}\right),$$from which $\triangle{ABQ}$ is determined by $\theta$. We optimize the expression above when $\theta$ is maximized, i.e. when $4\theta = 180^\circ \implies \theta = 45^\circ$. This happens when $Q$ lies on the line tangent to circle $\odot(O_2)$ at point $B$, as shown below.
[asy]
/* Made by MRENTHUSIASM */
size(200);
pair A, B, O1, O2, P, Q;
A = (-30, -10);
B = (0, 12);
O1 = (0, 0);
O2 = (0, 8);
P = (-11.174, 10.408);
Q = (-11.174, -8.408);
dot("$A$", A, W);
dot("$B$", B, dir(75));
dot("$O_2$", O2, W);
clip(anglemark(Q, B, A, 80));
draw(arc(O1, 12, 0, 180));
draw(arc(O2, 6, 0, 180));
draw(A--B);
draw(O2--B, gray);
draw(arc(B, 5, 95, 147));
draw(P--Q, gray);
[/asy]
Now, from $\triangle{ABQ} \sim \triangle{PBQ}$, we get
$$\frac{\overline{AB}}{6} = \frac{2\overline{BQ}}{12} = \frac{\overarc{BPQ}}{180^\circ} = \frac{4\theta}{180^\circ} = \frac{4(45^\circ)}{180^\circ} = \frac{1}{2} \implies \overline{AB} = \boxed{6}.$$
Let the centers of the circles be $O_1$ and $O_2$, and let $R_1$ and $R_2$ be the radii of the circles, with $R_1 = 6$ and $R_2 = 12$. Let $A$ and $B$ be the points of intersection of $\overline{AB}$ with the circles, with $A$ on the larger circle centered at $O_2$. The length of $\overline{AB}$ is given by $AB = AP + PB$.
Since $PQ = 8$, each of $AP$ and $PB$ is half of $PQ$, which means $AP = 4$ and $PB = 4$.
[asy]
unitsize(0.8 cm);
pair[] O;
real[] R;
pair P, Q;
R[1] = 6;
R[2] = 12;
O[1] = (-1,0);
O[2] = (1,0);
Q = intersectionpoint(arc(O[1],R[1],0,180),arc(O[2],R[2],0,180));
P = intersectionpoint(arc(O[1],R[1],0,180),arc(O[2],R[2],180,360));
dot("$P$", P, NW);
dot("$Q$", Q, SE);
draw(Circle(O[1],R[1]));
draw(Circle(O[2],R[2]));
draw((-R[2],0)--(R[2],0),red+1.5);
pair A = intersectionpoint(arc(O[2],R[2],180,225),(-R[2],0)--(R[2],0));
pair B = intersectionpoint(arc(O[1],R[1],45,180),(-R[2],0)--(R[2],0));
dot("$A$", A, NW);
dot("$B$", B, SE);
draw(A--P--B,dashed);
draw(anglemark(O[2],A,P,25));
draw(anglemark(P,B,O[1],25));
draw(anglemark(O[1],P,A,25));
draw(anglemark(B,Q,O[2],25));
label("$R_2 = 12$", O[2], S);
label("$R_1 = 6$", O[1], S);
[/asy]
In order to maximize the length of $\overline{AB}$, we want the segment $\overline{AB}$ to be a diameter of one of the circles. Let's consider the case when $\overline{AB}$ is a diameter of the larger circle centered at $O_2$.
Since $A$ is the endpoint of the diameter, we have $AO_2 = R_2 = 12$. Similarly, $O_1P = R_1 = 6$. Therefore, $AP = AO_2 - O_1P$. Using the fact that $AP = 4$, we have $4 = 12 - 6$, so $O_1P = 6$.
Now, we can apply the Pythagorean Theorem to the right triangle $\triangle O_1PB$ to solve for $PB$:
\[PB^2 = O_1P^2 + O_1B^2.\]Since $O_1B = R_1 = 6$, we have
\[PB^2 = 6^2 + 6^2 = 72.\]Therefore, $PB = \sqrt{72} = 6\sqrt{2}$.
Hence, the length of $\overline{AB}$ is
\[AB = AP + PB = 4 + 6\sqrt{2}.\]Therefore, the largest possible length of $\overline{AB}$ is $\boxed{4 + 6\sqrt{2}}$.