Two circles of radii 10 cm and 17 cm intersect at two points and the distance between their centres is 21 cm. Find the length of the common chord.

Asked by  | 30th Jan, 2012, 08:54: PM

Expert Answer:

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

Answered by  | 31st Jan, 2012, 09:55: AM

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