Here, we draw two circles `C_(1)` and `C_(2)` of radii `r_(1)=3cm` and `r_(2)=5cm` Now, we draw a chord AB such that it touches the circle `C_(1)` at point D. The centre of concentric circle is O.
Now, we draw a perpendicular bisector from O to AB which meets AB at D.
i.e.,`" "AD=BD`
In right `triangleOBD`
`OB^(2)=OD^(2)+DB^(2)" "`(by Pythagoras theorem)
`implies" "5^(2)=3^(2)+DB^(2)`
`DB^(2)=25-9=16`
`implies" "DB=4 cm`
Length of chord `= AB = 2BD = 2xx4 = 8 cm`