If z=x+i y satisfies |z|-2=0 and |z-i|-|z+5 i|=0, then :

Solution

Maths sir · 4 steps

Two conditions, two curves: find where the circle meets the bisector, then test each option there.

1. First condition. |z| = 2 means z is at distance 2 from the origin. So x² + y² = 4, a circle.
2. Second condition. |z - i| = |z + 5i| means z is equally far from i and -5i. Squaring gives x² + (y-1)² = x² + (y+5)², so -2y + 1 = 10y + 25, which gives y = -2.
3. Find the point. Put y = -2 in x² + y² = 4. Then x² = 0, so x = 0. The only such z is -2i, the point (0, -2).

   Option	Value at (0, -2)	Holds?
   (a) x + 2y - 4	0 - 4 - 4 = -8	no
   (b) x² + y - 4	0 - 2 - 4 = -6	no
   (c) x + 2y + 4	0 - 4 + 4 = 0	yes
   (d) x² - y + 3	0 + 2 + 3 = 5	no

4. So. Only x + 2y + 4 = 0 is satisfied by z = -2i, so the answer is (c).

The idea to keep. |z - a| = |z - b| is the perpendicular bisector of a and b. When the conditions pin z to a single point, substitute that point into every option.

Answer from the NTA final answer key. Solution by PrepWiser.