Question

PLS ANSWER QUICKLY TOMORROW EXAM
two circles of radius 5 cm and 3 cm intersect at two points and the distance between their centers is 4 cm find the length of the common chord

Answer 1

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.