Given: z z̅ + (1 + i) z +(1 - i) z̅ = 0 As we know, z = x + iy, z̅= 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.