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Answer

`(-(9)/(5),(12)/(5))and((9)/(5),-(12)/(5))`

Explanation

Given,
and let `C_(2):(x-h)^(2)+(y-k)^(2)=25`
`therefore` Equation of common chords is `S_(1)-S_(2)=0`
`therefore 2hx+2ky=(h^(2)+k^(2)-9)`
If p be the length of perpendicular on it from the centre
(0, 0) of `C_(1)` of radius 4, then `p=(h^(2)+k^(2)-9)/(sqrt(4h^(2)+4k^(2)))`.
Also, the length of the chord is
`2sqrt(r^(2)-p^(2))=2sqrt(4^(2)-p^(2))`
The chord will be of maximum length, if ` phi = 0` or
`h^(2)+k^(2)-9=0rArr+(16)/(9)h^(2)=9`
`rArr h=pm(9)/(5)`
`therefore k=pm(12)/(5)` Hence,centres ar `((9)/(5),(12)/(5)) and (-(9)/(5), (12)/(5))`