If Rez−12z+i=1, where z = x + iy, then the point (x, y) lies on a

Let z=x+iy.  Thenz−12z+i=x−1+iy2x+iy+i=x−1+iy2x+2y−1i×2x−2y+1i2x−2y+1Now Rez+12z+i=2xx−1+y2y+12x2+2y+12=1⇒2x2+2y2−2x+y=4x2+4y2+4y+1⇒2x2+2y2+2x+3y+1=0⇒x2+y2+x+32y+12=0circle with centre −12,−34r=14+916−12=4+9−816=54