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 $\overline{AB}$.

Answer 1

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
$$\frac{\overline{AB}}{6} = \frac{2\overline{BQ}}{12} = \frac{\overarc{BPQ}}{180^\circ} = \frac{4\theta}{180^\circ} = \frac{4(45^\circ)}{180^\circ} = \frac{1}{2} \implies \overline{AB} = \boxed{6}.$$