Two circles touch externally. The sum of their areas is 130π cm2 and the distance between their centers is 14 cm. The radius of the smaller circle is-

Let radii be r and R. For external tangency: r + R = 14 cm. Sum of areas: πr² + πR² = 130π, so r² + R² = 130. Squaring the first equation: (r+R)² = r² + 2rR + R² = 196. Substituting: 130 + 2rR = 196, so rR = 33. Solving the quadratic x² - 14x + 33 = 0 gives (x-3)(x-11) = 0, so radii are 3 cm and 11 cm. The smaller radius is 3 cm.