two circles of radii 5cm and 3cm intersect at two points and the distance between their centre is 4 cm find the length of common chord
Answers
Step-by-step explanation:
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➧ Let the radius of the two circles be 5 cm and 3 cm respectively whose centre’s are O and O'.
➧ Hence OA = OB = 5 cm
➾ O'A = O'B
➾ 3 cm
➧ OO' is the perpendicular bisector of chord AB.
➧ Therefore, AC = BC
➧ Given, OO' = 4 cm
➧ Let OC = x
➧ Hence O'C = 4 − x
➧ In right angled ΔOAC, by Pythagoras theorem OA²
➾ OC² + AC²
➾ 5² = x² + AC²
➾ AC² = 25 − x²à
➧ (1) In right angled ΔO'AC, by Pythagoras theorem O'A²
➾ AC² + O'C²
➾ 3² = AC² + (4 – x)²
➾ 9 = AC² + 16 + x² − 8x
➾ AC² = 8x − x² − 7 à
➧ (2) From (1) and (2),
➧ we get 25 − x² = 8x − x² − 7 8x = 32
➧ Therefore, x = 4
➧ Hence, the common chord will pass through the centre of the smaller circle, O' and hence, it will be the diameter of the smaller circle.
➧ AC² = 25 − x²
➾ 25 − 4²
➾ 25 − 16 = 9
➧ Therefore, AC = 3 m Length of the common chord,
➾ AB = 2AC
➾ 6 cm ...✔
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