If `z_1`, lies in `|z-3|<=4,z_2` on `|z-1|+|z+1|=3 and A = |z_1-z_2|`, then :

Options

A. `0 le A le 15/2`
B. `0 lt A lt 15/2`
C. `0 le A le 17/2`
D. `0 le A lt 17/2`

Answer

D

Explanation

We observe that `|z-3| le 4` represent the set of all interior and boundary points of the circle having center at `(3,0)` and radius 4.
The equation `|z-1|+|z+1|=3` represents the set of all points lying on the boundary of the ellipse having its two foci at `(-1,0)` and `(1,0)` and major axis of length 3.

Clearly, `A=|z_(1)-z_(2)|=0` when `z_(1)=z_(2)` (=affix of P or Q)
`therefore "min".|z_(1)-z_(2)|=0`
Also, `A=|z_(1)-z_(2)|`is greatest when `z_(1)` = affix of B and `z_(2)`= affix of a.
`therefore` max. `|z_(1)-z_(2)|=A^(')B=A^(')O+OB+3/2+7=17/2`
Hence, `0 le A le 17/2`.