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.