Question

Two circles of radii 5 cm and 3 cm intersect at two points and the distance between
their centres is 4 cm. Find the length of the common chord

Answer 1

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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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Answer 2

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 .

Hope it helped you.