Question
Question: If there exists an Z satisfying both \|Z –mi\| = m + 5 and \|Z–4\| \< 3, then the set of all permiss...
If there exists an Z satisfying both |Z –mi| = m + 5 and |Z–4| < 3, then the set of all permissible values of m belong to the set–
A
(–3, 3)
B
(–3, 9)
C
(–5, –3)
D
(4, 9)
Answer
(–3, 3)
Explanation
Solution
Sol. |Z –mi| = m + 5 represents a circle with mi or B(0, m) as centre and radius m + 5.
|Z –4| < 3 represents the interior of a circle with centre 4 or A(4, 0) and radius 3.
If there is to be at least one Z satisfying both the two circles should intersect.
(i.e.) r1~ r2 < d < r1 + r2
m + 5 – 3 < m2+16< m + 5 + 3
squaring, m2 + 4m + 4 < m2 + 16 < m2 + 16 m + 64
\ m < 3 and m > –3
\ m Ī (–3, 3)