ěŹ.to Raio, and Ha distante tekueeth t.list te

1.	ěŹ.to Raio, and Ha distante tekueeth t.list te
Answer» 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 theoremOA² ➾ OC²+ AC² ➾ 5²= x² + AC² ➾ AC² = 25 −x²à ➧ (1)In right angled ΔO'AC, by Pythagoras theoremO'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 get25 −x² = 8x −x²− 78x= 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 m ...✔

Answer»

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 theoremOA²

➾ OC²+ AC²

➾ 5²= x² + AC²

➾ AC² = 25 −x²à

➧ (1)In right angled ΔO'AC, by Pythagoras theoremO'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 get25 −x² = 8x −x²− 78x= 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 m ...✔

Discussion