Correct Answer - C::D
Radius of director circle `sqrt(a^(2)-b^(2)) =2`
Axis of hyperbola is line joining the center `C(0,2)` and `A(2,0)` (end of major axis)
`:. a = CA = 2 sqrt(2)`
`:. (2sqrt(2))^(2)-b^(2)=4`
`:. b^(2) =4`
`:. e = (sqrt(3))/(sqrt(2))`
Center of the hyperbola is center of the director circle (0,2) Focus lies on this line at distance ae from center
`:.` If foci are (x,y) then `(x-0)/(-(1)/(sqrt(2))) = (y-2)/((1)/(sqrt(2))) = +- 2 sqrt(3)`
`(x,y) = (sqrt(6),2-sqrt(6))` or `(-sqrt(6),2+sqrt(6))`
