[asy]
size(150);
dotfactor=4;
draw(circle((0,0),4));
draw(circle((10,-6),3));
pair O,A,P,Q;
O = (0,0);
A = (10,-6);
P = (-.55, -4.12);
Q = (10.7, -2.86);
dot("O", O, NE);
dot("O′", A, SW);
dot("P", P, SW);
dot("Q", Q, NE);
draw((2*sqrt(2),2*sqrt(2))--(10 + 3*sqrt(2)/2, -6 + 3*sqrt(2)/2)--cycle);
draw((-1.68*sqrt(2),-2.302*sqrt(2))--(10 - 2.6*sqrt(2)/2, -6 - 3.4*sqrt(2)/2)--cycle);
draw(P--Q--cycle);
//Credit to happiface for the diagram[/asy]In the adjoining figure, every point of circle O′ is exterior to circle O. Let P and Q be the points of intersection of an internal common tangent with the two external common tangents. Then the length of PQ is<spanclass=′latex−bold′>(A)</span>the average of the lengths of the internal and external common tangents<spanclass=′latex−bold′>(B)</span>equal to the length of an external common tangent if and only if circles O and O′ have equal radii<spanclass=′latex−bold′>(C)</span>always equal to the length of an external common tangent<spanclass=′latex−bold′>(D)</span>greater than the length of an external common tangent<spanclass=′latex−bold′>(E)</span>the geometric mean of the lengths of the internal and external common tangents