Answer :


Let Center of the Circles are A and B. CD is a common Chord of the circle. AB is the perpendicular bisector of the chord CD.


If AB is a perpendicular bisector of CD then it should passes through both the centers.


So, AB is the distance that we need to calculate.


Given, AC = 10cm, CD = 12cm


CM


CM


CM = 6cm


In ACM


AC2 = AM2 + CM2


AM2 = AC2-CM2


AM2 = 100-36


AM2 = 64


AM = 8cm


In BCM


BC2 = BM2 + CM2


BM2 = BC2-CM2


BM2 = 100-36


BM2 = 64


BM = 8cm


AB = 16cm


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