UPSC NDA Math Model Paper 2

Given:
zz+(1+i)z+(1−i)z=0
As we know,
z=x+iy,z=x− iy using these value in the above given equation we get:
(x+iy)(x−iy)+(1+i)(x+iy) +(1−i)(x−iy)=0
x2+y2+x+iy+ix−y+x−iy −ix−y=0
x2+y2+2x−2y=0 -----(3)
Comparing (3) with (1) we get:
h = -1 and k = 1 which are centre of circle.
By using h, k values in (2) we get the radius of the circle as √2.