Given: Two circles have the same center O and AB is a chord of the larger circle touching
the smaller circle at C. also, OA = 5 cm ad OC = 3 cm
In △OAC, OA2 = OC2 + AC2
∴ AC2 = OA2 - OC2
\(\Rightarrow\) AC2 = 52 - 32
\(\Rightarrow\) AC2 = 25 - 9
\(\Rightarrow\) AC2 = 16
\(\Rightarrow\) AC = 4cm
∴ AB = 2AC
(Since perpendicular drawn from the center of the circle bisects the chord)
∴ AB = 2 x 4 = 8cm
The length of the chord of the larger circle is 8cm.