The two circles below are externally tangent. A common external tangent intersects line at Find line PQ at R. The radius of (P) is 12, and the radius of (Q) is 8. Find QR.
Let $O_1$ and $O_2$ be the centers of circles $P$ and $Q$ respectively. Let $A$ and $B$ be the points where line $PQ$ intersects the circles $P$ and $Q$ respectively. Let $C$ be the midpoint of $\overline{AB}$. Let $D$ be the foot of the perpendicular from $O_2$ to $\overline{PQ}$, and let $E$ be the foot of the perpendicular from $O_1$ to $\overline{O_2D}$.
[asy]
pair O1, O2, A, D, E, P, Q;
O1=(0,0);
O2=(20,0);
P=O1--(0, 12);
Q=O2--(0, 8);
D=foot(O2, P, Q);
E=foot(O1, O2, D);
A=IP(P, D--(100, 4500/7));
draw(Circle(O1, 12));
draw(Circle(O2, 8));
draw(P--O2--D--O1);
draw(O2--E);
draw(P--Q);
label("12", O1--A, W);
label("8", O2--B, E);
label("A", A, W);
label("B", B, E);
label("C", (A+B)/2, N);
label("P", P, NW);
label("Q", Q, NE);
label("O2", O2, S);
label("D", D, S);
label("O1", O1, SW);
label("E", E, S);
[/asy]
Since $\overline{O_1E}$ and $\overline{O_2D}$ are both perpendicular to $\overline{PQ}$, $O_1EDO_2$ is a rectangle. Therefore, $O_1E=O_2D$. Since $O_1O_2=12+8=20$, we know $O_1E=O_2D=10$. Because $\angle O_1AQ=\angle O_2BQ=90^\circ$, angles $O_1EA$ and $O_2DE$ are both right angles. Thus, by the Pythagorean Theorem on $\triangle O_1AE$, \[AE=\sqrt{12^2-10^2}=2\sqrt{7}\]and by the Pythagorean Theorem on $\triangle O_2DE$, \[ED=\sqrt{20^2-10^2}=10\sqrt3.\]
Finally, we find the length of $\overline{QR}$. By the Pythagorean theorem on $\triangle CDE$, \[QR = CD = \sqrt{AE^2+DE^2}=\sqrt{(2\sqrt7)^2+(10\sqrt3)^2} = \boxed{4\sqrt{10}}.\]