putting in z= x+iy and simplifying we get
x= 1+b2-a22(b-a)
thus locus should be Re(z) = 1+b2-a22(b-a)
\hspace{-16}$If $\mathbf{a,b\in \mathbb{R}}$ and $\mathbf{a\neq b}$. Then Locus of all Complex no. $\mathbf{z}$ which\\\\ Satisfy the equation. $\mathbf{\mid z-a \mid^2-\mid z-b \mid^2=1}$
putting in z= x+iy and simplifying we get
x= 1+b2-a22(b-a)
thus locus should be Re(z) = 1+b2-a22(b-a)