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

\[ \begin{aligned} & \text { Centre } \equiv\left(-\frac{\mathrm{k}_{1}}{2}, \frac{\mathrm{k}_{2}}{2}\right) \equiv(1,2) \Rightarrow \mathrm{k}_{1}=-2, \mathrm{k}_{2}=4 \text { radius }=1 \Rightarrow \lambda=4 \\ & \operatorname{Im}(\mathrm{w})=\mathrm{k}_{1} \mathrm{y}+\mathrm{k}_{2} \mathrm{x}+\lambda=0 \therefore \text { line } \mathrm{AB}, 2 \mathrm{x}-\mathrm{y}+2=0 \\ & \mathrm{~d}=\left|\frac{2-2+2}{\sqrt{4+1}}\right|=\frac{2}{\sqrt{5}} \text { let } \mathrm{AB}=\ell, \mathrm{AM}^{2}=\mathrm{AC}^{2}-\mathrm{d}^{2} \\ & \frac{\ell^{2}}{4}=1-\frac{4}{5}=\frac{1}{5} \therefore 30 \ell^{2}=24 \end{aligned} \]