If |2z−4−2i|=|z|sin⁡π4−arg⁡z , then locus of z is/an

Let z=x+iy=r(cos⁡θ+isin⁡θ) ,  Then the equation is |2(x−2)+2i(y−1)|=r12cos⁡θ−12sin⁡θ=12(rcos⁡θ−rsin⁡θ)⇒(x−2)2+(y−1)2=12x−y2 It is an ellipse with focus at (2,1) and directrix x−y=0 and eccentricity =12<1 The locus represents an ellipseTherefore, the correct answer is (1).