Two circles centered at A and B, each with a radius of 10 cm, intersect at points P and Q. PQ = 16 cm AB intersects PQ at point O. \(\therefore\) In triangle AOP, AP = 10 cm OP = \(\frac{PQ}{2} = \frac{16}{2}\) cm = 8 cm \(\therefore\) AO = \(\sqrt{AP^2 - OP^2} = \sqrt{10^2 - 8^2}\) cm \(= \sqrt{100 - 64} = \sqrt{36} = 6\) cm \(\therefore\) AB = 2AO = \(2 \times 6\) cm = 12 cm \(\therefore\) The distance between the centers of the two circles is 12 cm.