Question

Let ω = z\(\bar{z}\) + k₁z + k₂iz + λ(1+i), k₁ , k₂ ∈ \(\mathbb{R}\). Let Re(ω) = 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 (ω) = 0 intersects C at A and B, then 30 (AB)² is equal to _____

Answer 1

Correct Answer: 24

The given expression is \[ \omega = z\bar{z} + k_1z + k_2\bar{z} + \lambda (1+i). \] The real part of \(\omega\) is: \[ \text{Re}(\omega) = x^2 + y^2 + k_1x + k_2y + \lambda = 0. \]

Step 1: Find the center and radius of the circle

Comparing this with the equation of a circle, \[ x^2 + y^2 + 2gx + 2fy + c = 0, \] we find the center as \[ \text{Center} = \left(-\frac{k_1}{2}, -\frac{k_2}{2} \right). \] It is given that the circle \(C\) is in the first quadrant, touches \(y = 1\) and the \(y\)-axis, and has a radius of 1. Hence, \[ -\frac{k_1}{2} = 1, \quad -\frac{k_2}{2} = 2, \quad \lambda = 4. \] Thus, \[ k_1 = -2, \quad k_2 = -4, \quad \lambda = 4. \] The center of the circle is \((1, 2)\), and the radius is 1.

Step 2: Curve Im(\(\omega\)) = 0

The imaginary part of \(\omega\) is: \[ \text{Im}(\omega) = k_1y - k_2x + \lambda = 0. \] Substitute \(k_1 = -2, k_2 = -4, \lambda = 4\): \[ -2y - (-4)x + 4 = 0 \quad \Rightarrow \quad -2y + 4x + 4 = 0 \quad \Rightarrow \quad 2x - y + 2 = 0. \]

Step 3: Intersection points \(A\) and \(B\)

The line \(2x - y + 2 = 0\) intersects the circle \(x^2 + y^2 + 2x - 4y + 4 = 0\). Solving these equations gives the points of intersection \(A\) and \(B\). From \(y = 2x+2\), substitute into the circle equation: \[ x^2 + (2x+2)^2 + 2x - 4(2x+2) + 4 = 0 \] \[ x^2 + 4x^2 + 8x + 4 + 2x - 8x - 8 + 4 = 0 \] \[ 5x^2 + 2x = 0 \] \[ x(5x+2) = 0 \] So \(x=0\) or \(x=-\frac{2}{5}\). If \(x=0\), \(y=2\). If \(x=-\frac{2}{5}\), \(y=2(-\frac{2}{5})+2 = \frac{6}{5}\). So \(A = (0,2)\) and \(B = (-\frac{2}{5}, \frac{6}{5})\). \[ AB = \sqrt{(0+\frac{2}{5})^2 + (2-\frac{6}{5})^2} = \sqrt{\frac{4}{25} + \frac{16}{25}} = \sqrt{\frac{20}{25}} = \frac{2\sqrt{5}}{5}. \]

Step 4: Calculate \(30(AB)^2\)

\[ (AB)^2 = \left(\frac{2\sqrt{5}}{5}\right)^2 = \frac{20}{25} = \frac{4}{5}. \] \[ 30(AB)^2 = 30 \cdot \frac{4}{5} = 24. \] Final Answer: \[ \boxed{24}. \]