Let w=z z̄+k₁ z+k₂ i z+λ(1+i), k₁, k₂ ∈ ℝ. Let Re(w)=0 be the circle C of radius 1 in the first quadrant touching the line y=1 and the y-axis. If the curve Im(w)=0 intersects C at A and B, then 30(A B)² is equal to __________

Solution

Maths sir · 5 steps

Split w into real and imaginary parts: one is the circle C, the other is a line cutting it.

1. Separate the parts. Put z = x + iy. Then zz̄ = x² + y², k₁ z = k₁ x + i k₁ y and k₂ i z = -k₂ y + i k₂ x. This gives Re(w) = x² + y² + k₁ x - k₂ y + λ and Im(w) = k₂ x + k₁ y + λ.
2. Find circle C. Touching the y-axis with radius 1 puts the centre at x = 1. Touching y = 1 puts it at y = 0 or y = 2, and only y = 2 keeps the circle in the first quadrant. So C is (x-1)² + (y-2)² = 1, that is, x² + y² - 2x - 4y + 4 = 0.
3. Match coefficients. Comparing with Re(w) = 0 gives k₁ = -2, k₂ = 4, λ = 4. So Im(w) = 0 is the line 4x - 2y + 4 = 0, or 2x - y + 2 = 0.
4. Chord length. The distance from the centre (1, 2) to the line is d = (|2 - 2 + 2|)/√5 = 2/√5. The half-chord satisfies (AB/2)² = r² - d² = 1 - 4/5 = 1/5.

   (AB)² = 4(r² - d²) = 4 × 1/5 = 4/5

5. So. 30(AB)² = 30 × 4/5 = 24.

The common slip: 6. Taking r² - d² = 1/5 as (AB)² forgets that it is the half-chord squared, and gives 6 instead of 24.

The idea to keep. When a complex expression is set to zero in its real or imaginary part, write z = x + iy and read off a curve in x, y. For a chord, use AB = 2√(r² - d²).

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