Let w = z z ¯ + k 1 z + k 2 i z + λ ( 1 + i ) , k 1 , k 2 ∈ ℝ . . Let R e ( w ) = 0 be…

Given,

$w=z\overline{z}+k_{1}z+k_{2}iz+\lambda (1+i),k_1,k_2\in \mathrm{ℝ}.$ .

And $Re\left(w\right)=0$,

So, taking $z=x+iy \& \overline{z}=x-iy$

We get,

$w=x^2+y^2+k_1\left(x+iy\right)+k_{2}i\left(x+iy\right)+\lambda (1+i)$

$\Rightarrow \mathit{Re}\left(\mathit{w}\right)=0\Rightarrow x^2+y^2+k_{1}x-k_{2}y+\lambda =0$

And, $\mathit{Im}\mathit{\left(}w\mathit{\right)}=0\Rightarrow k_{1}y+k_{2}x+\lambda =0$

Now given $Re\left(w\right)=0$ circle is touching $y=1$ and $y-$axis, with radius $1$, so we have equation of circle

$x^2+y^2-2x-4y+4=0$

So, on comparing with $Re\left(w\right)=0$ we get,

$k_1=-2,k_2=4,\lambda =4$

So, $\mathit{Im}\mathit{\left(}w\mathit{\right)}=0\Rightarrow -2y+4x+4=0$

And given $-2 y+4 x+4=0$ intersects the circle,

So, solving $x^2+y^2-2x-4y+4=0$ and $-2 y+4 x+4=0$ we get,

Intersecting point as $A\left(0,2\right)$ and $B\left(\frac{2}{5},\frac{14}{5}\right)$

Hence, by distance formula we get,

$30(AB{)}^2=30\left[{\left(0-\frac{2}{5}\right)}^2+{\left(2-\frac{14}{5}\right)}^2\right]=24$