Let $r_1$ be the radius of the inner circle and $r_2$ be the radius of the outer circle.From the problem, we know that $r_1 + r_2 = 16\,\text{cm}$The length of chord AB is 16 cm, so half of it (AP) is 8 cm.Using the Pythagorean theorem in triangle OAP,we have $r_2^2 = r_1^2 + 8^2$which simplifies to $r_2^2 = r_1^2 + 64$Now, substitute $r_2 = 16 - r_1$ into the equation $r_2^2 = r_1^2 + 64$and solve for $r_1$ and $r_2$.The difference of the radii is 8 cm.