Answer :
The perpendicular bisector of the common chord AB is the straight line joining the two centers O and O'. Let OO' intersect AB at the mid point of AB at C.
So ACO and ACO' are two right angle triangles. and we have OO' = OC+CO' = 4 cm. OA=5 cm and O'A = 3 cm.
AO² = OC² + CA²
5² = OC² + CA² --- (1)
O'A² = O'C² + CA²
3² = (OO'-OC)² + CA² = (4-OC)²+CA² =
9 = 16 - 8 OC + OC² + CA² ---- (2)
From (1) and (2) , we get:
9 = 16 - 8 * OC + 5² => OC = 4 cm
From (1), CA² = 5² - 4² => CA = 3 cm
Length of the common chord is twice CA = 6 cm.
So ACO and ACO' are two right angle triangles. and we have OO' = OC+CO' = 4 cm. OA=5 cm and O'A = 3 cm.
AO² = OC² + CA²
5² = OC² + CA² --- (1)
O'A² = O'C² + CA²
3² = (OO'-OC)² + CA² = (4-OC)²+CA² =
9 = 16 - 8 OC + OC² + CA² ---- (2)
From (1) and (2) , we get:
9 = 16 - 8 * OC + 5² => OC = 4 cm
From (1), CA² = 5² - 4² => CA = 3 cm
Length of the common chord is twice CA = 6 cm.