Correct Answer - A
Let the equation of the circle be `x^(2)+y^(2)+2gx+2fy+c=0`. It touches x-axis at a distance 3 from the origin. Therefore, `c=g^(2)` and the circle passes through `(pm3, 0)`.
`:. 9 pm 6g +c=0`
`rArr 9 pm 6g + g^(2)= 0 rArr (gpm 3)^(2)=0 rArr g=-3`
`:. c=g^(2)rArr c=9`
The circle cuts an intercept of length `2 sqrt(17)` on y-axis.
`:. 2 sqrt(f^(2)-c)=2sqrt(7)rArr f^(2)-9=7rArr f = pm 4`
Hence, the equations of the circles are
`x^(2)+y^(2)-6xpm 8y+9=0`