CBSE Class 9 Answered

Let the radius of the circle centred at O and O' be 17 cm and 10 cm respectively. OA = OB = 17 cm O'A = O'B = 10 cm OO' will be the perpendicular bisector of chord AB. ? AC = CB It is given that, OO' = 21 cm Let OC be x. Therefore, O'C will be 21 ? x. In ?OAC, OA2 = AC2 + OC2 ? 172 = AC2 + x2 ? 289 – x2 = AC2 ... (1) In ?O'AC, O'A2 = AC2 + O'C2 ? 102 = AC2 + (21 ? x)2 ? 100 = AC2 + 441 + x2 ? 42x ? AC2 = ? x2 - 341 + 42x ... (2) From equations (1) and (2), we obtain 289? x2 = ? x2 ? 341 + 42x 42x = 630 x = 15 Now, AC2 = 289 – x2 = 289 ? 152 = 289 ? 225 = 64 ? AC = 64 cm Length of the common chord AB = 2 AC = (2 × 64) cm = 128 cm