Let C be the circle in the complex plane with centre z₀=1/2(1+3 i) and radius r=1. Let z₁=1+i and the complex number z₂ be outside the circle C such that |z₁-z₀||z₂-z₀|=1. If z₀, z₁ and z₂ are collinear, then the smaller value of |z₂|² is equal to :

Solution

Maths sir · 5 steps

Move along the line from z₀ through z₁ a distance √2 each way, then pick the closer point.

1. Distance to z1. z₁ - z₀ = (1 + i) - (1/2 + 3/2i) = 1/2 - 1/2i, so |z₁ - z₀| = 1/√2.
2. Distance to z2. The condition |z₁ - z₀| |z₂ - z₀| = 1 gives |z₂ - z₀| = √2. This is more than the radius 1, so z₂ is outside C as required.
3. Use collinearity. z₂ lies on the line through z₀ and z₁. The unit vector along it is (z₁ - z₀)/(|z₁ - z₀|) = (1 - i)/√2. Going √2 in either direction adds or subtracts 1 - i.

   z₂ = z₀ ± (1 - i) = 3/2 + 1/2i or -1/2 + 5/2i

4. Find the moduli. For 3/2 + 1/2i: |z₂|² = 9/4 + 1/4 = 5/2. For -1/2 + 5/2i: |z₂|² = 1/4 + 25/4 = 13/2.
5. So. The smaller value of |z₂|² is 5/2, which is option (b).

The idea to keep. A point at a given distance from z₀ on a known line is z₀ plus or minus that distance times the unit vector along the line. Always check both directions.

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