Two circles O and Q cross each other at points A and B. Connect those 4 points O, Q, A and B to each other. Here OA=OB=QA=QB=10 ( given radii ), so quadrilateral OAQB is a rhombus. Therfore 2 diagonals, AB and OQ, perpendicularly bisect each other at the midpoint of AB: H. So HA=HB=AB/2=6cm ( AB: given length ), OH=QH and ∠OHA=90°. Thus OQ=OH+QH=2xOH, and △OHA is a right triangle, so OA^2=OH^2+HA^2 ( the Pythagorean theorem ). Plug OA=10 and HA=6 into the equation above, and find the value of OH. 10^2=OH^2+6^2, OH^2=10^2-6^2=100-36=64, so OH=8. Therefore OQ=2xOH=2x8=16. The length of OQ is 16.