JEE Main 2023 — Mathematics Algebra

Given,

$w=z\bar{z}+{k}_{1}z+{k}_{2}iz+\lambda (1+i),{k}_{1},{k}_{2}\in \mathbb{R}.$w=zzˉ+k1​z+k2​iz+λ(1+i),k1​,k2​∈R. .

And $Re(w)=0$Re(w)=0,

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

We get,

$$w={x}^{2}+{y}^{2}+{k}_{1}(x+iy)+{k}_{2}i(x+iy)+\lambda (1+i)$$w=x2+y2+k1​(x+iy)+k2​i(x+iy)+λ(1+i)

$$\Rightarrow Re(w)=0\Rightarrow {x}^{2}+{y}^{2}+{k}_{1}x-{k}_{2}y+\lambda =0$$⇒Re(w)=0⇒x2+y2+k1​x−k2​y+λ=0

And, $Im(w)=0\Rightarrow {k}_{1}y+{k}_{2}x+\lambda =0$Im(w)=0⇒k1​y+k2​x+λ=0

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

$${x}^{2}+{y}^{2}-2x-4y+4=0$$x2+y2−2x−4y+4=0

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

$${k}_{1}=-2,{k}_{2}=4,\lambda =4$$k1​=−2,k2​=4,λ=4

So, $Im(w)=0\Rightarrow -2y+4x+4=0$Im(w)=0⇒−2y+4x+4=0

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

So, solving ${x}^{2}+{y}^{2}-2x-4y+4=0$x2+y2−2x−4y+4=0 and $-2y+4x+4=0$−2y+4x+4=0 we get,

Intersecting point as $A(0,2)$A(0,2) and $B(\frac{2}{5},\frac{14}{5})$B(52​,514​)

Hence, by distance formula we get,

$$30(AB{)}^{2}=30[{(0-\frac{2}{5})}^{2}+{(2-\frac{14}{5})}^{2}]=24$$30(AB)2=30[(0−52​)2+(2−514​)2]=24