TS EAMCET 11-Sep-2020 Shift 2 Solved Paper

Circle ⇒|z|2+2⋅Re(az)−4=0.....(i) [z+z=2Re(z)]
Now, given z=x+iy⇒|z|2=x2+y2
and a=1+i, then az=(1+i)(x+iy)
∴ From Eq. (i), put Re(az)=x−y
and ‌‌|z|2=x2+y2
⇒‌‌x2+y2+2(x−y)−4=0
⇒S:x2+y2+2x−2y−4=0...(i)
is a circle and given line
L:z+z−i(z−z)+2=0
⇒‌‌L:2Re(z)−i2Im(z)+2=0
⇒‌‌L:2x−i(2yi)+2=0
⇒‌‌L:x+y+1=0...(iii)
Now, equation of circle passing through point of intersection of circle S and line L is
S+λL=0
Pass through (0,0)
∴ Put in Eq. (iv), we get
−4+λ=0⇒λ=4
Now, required circle is
⇒‌‌x2+y2+6x+2y=0