JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Complex Numbers

Let $z=x+iy$. Then $z\overline{z}=x^2+y^2$, $k_{1}z=k_1(x+iy)$, $k_{2}iz=k_{2}i(x+iy)=k_2(ix-y)$. So $\omega =x^2+y^2+k_{1}x+k_{2}ix-k_{2}y+iy\lambda +\lambda$. Thus $\mathrm{Re}(\omega )=x^2+y^2+k_{1}x-k_{2}y+\lambda =0$. This is a circle with center $(-k_1/2,k_2/2)$. Given it has radius 1, touches $y=1$ and the y-axis in the first quadrant, the center is $(1,2)$. So $k_1=-2$, $k_2=4$, and $\lambda =4$. The circle is $x^2+y^2-2x-4y+4=0$. Now $\mathrm{Im}(\omega )=k_{1}y+k_{2}x+\lambda =-2y+4x+4=0$, i.e., $2x-y+2=0$. The distance from center $(1,2)$ to this line is $d=\frac{|2(1)-2+2|}{\sqrt{5}}=\frac{2}{\sqrt{5}}$. The chord length is $AB=2\sqrt{r^2-d^2}=2\sqrt{1-4/5}=2\sqrt{1/5}=\frac{2}{\sqrt{5}}$. Thus $(AB{)}^2=4/5$, and $30(AB{)}^2=30\cdot 4/5=24$.