Q.If $z=x+iy$ and $|z-zi|=1$ then : (A) $z$ lies on X-axis (B) $z$ lies on Y-axis (C) $z$ lies on a circle (D) $z$ lies on a rectangle

If $z=x+iy$ represents a variable point $P(x,y)$ and $z_1=x_1+iy_1$ represents a fixed point $A(x_1,y_1)$ in the Argand plane, then $|z-z_1|$ is precisely the ordinary Euclidean distance between $P$ and $A$, computed by the distance formula $\sqrt{(x-x_1)^2+(y-y_1)^2}$. This single geometric fact converts modulus conditions on $z$ directly into familiar Cartesian curves. If $|z-z_1|=a$ for a fixed positive constant $a$, every point $z$ satisfying it sits at the fixed distance $a$ from $z_1$, so the locus is a circle centred at $z_1$ with radius $a$ — squaring both sides gives the Cartesian equation $(x-x_1)^2+(y-y_1)^2=a^2$ directly. If instead $|z-z_1|=|z-z_2|$ for two fixed points $z_1,z_2$, every point $z$ is equidistant from both, so the locus is the perpendicular bisector of the segment joining $z_1$ and $z_2$ — expanding both sides of $(x-x_1)^2+(y-y_1)^2=(x-x_2)^2+(y-y_2)^2$ and cancelling the squared terms leaves a linear equation, i.e. a straight line. These two cases (circle and perpendicular bisector) are the two standard locus types built from modulus conditions, and they are proved by translating the modulus/distance statement into coordinates and simplifying algebraically, exactly …