Let A = \z C : |z - 2| 4\ and B = \z C : |z - 2| + |z + 2| = 5\. Then the \|z1 - z2| : z1 A and z2 B\ is :
- 172
- 8
- 9
- 152
Answer
(1) 172
Solution
A is the closed disc with centre S(2, 0) and radius 4.
B: |z - 2| + |z + 2| = 5 > 4, so B is an ellipse with foci ( 2, 0), 2a = 5 a = 52, ae = 2.
For a fixed z2 B, the farthest point of the disc A is at distance |z2 - 2| + 4, so |z1 - z2| = 4 + |z2 - 2|.
|z2 - 2| is the focal distance from S(2, 0), SP = a - ex, which is largest at x = -a:
|z2 - 2| = a + ae = 52 + 2 = 92 (at z2 = - 52)
Required maximum = 4 + 92 = 172
Answer: (1)