Let $w=z \bar{z}+k_{1} z+k_{2} i z+\lambda(1+i), k_{1}, k_{2} \in \mathbb{R}$. Let $\operatorname{Re}(w)=0$ be the circle $\mathrm{C}$ of radius 1 in the first quadrant touching the line $y=1$ and the $y$-axis. If the curve $\operatorname{Im}(w)=0$ intersects $\mathrm{C}$ at $\mathrm{A}$ and $\mathrm{B}$, then $30(A B)^{2}$ is equal to __________

Given the expression for $w$ as : <br/><br/>$$w = z\bar{z} + k_1z + k_2iz + \lambda(1+i), \quad \text{where } k_1, k_2 \in \mathbb{R}.$$ <br/><br/>1. If $w = x+iy$, we can separate this into the real and imaginary parts : <br/><br/> The real part is: $\text{Re}(w) = x^2 + y^2 + k_1x - k_2y + \lambda = 0.$ <br/><br/> The imaginary part is: $\text{Im}(w) = k_1y + k_2x + \lambda = 0.$ <br/><br/>2. We know the circle C of radius 1 touches the line $y=1$ and the y-axis. The standard form of the equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center of the circle and $r$ is the radius. If the circle touches y-axis, then the x-coordinate of the center $h$ must be equal to the radius. Also, the circle touches the line $y=1$, so the y-coordinate of the center $k$ must be 1 unit above this line, hence $k=2$. So the circle's equation is $(x-1)^2 + (y-2)^2 = 1$. <br/><br/>3. By comparing this with $\text{Re}(w) = 0$, we can see that $k_1=-2$, $k_2=4$, and $\lambda=4$. <br/><br/>4. The equation for the curve $\text{Im}(w)=0$ becomes $-2y + 4x + 4 = 0$, which simplifies to $y = 2x + 2$. <br/><br/>5. To find the intersection points A and B, we solve the system of equations consisting of the circle and the line. Substituting $y = 2x + 2$ into the equation of the circle gives $5x^2 - 2x = 0$, with solutions $x = 0$ and $x = 2/5$. <br/><br/>6. Substituting $x = 0$ and $x = 2/5$ into $y = 2x + 2$ gives $y = 2$ and $y = 6/5$ respectively. So, the intersection points are A = $(0, 2)$ and B = $(2/5, 14/5)$. <br/><br/>7. The distance between A and B is $AB = \sqrt{[(2/5)^2 + (4/5)^2]} = \sqrt{4/5}$. <br/><br/>8. Finally, $30(AB)^2 = 30 \times (4/5) = 24$.