Question

The equation $|z-z_0|=r$ represents

(Given : $z_0$ is a fixed complex number.)

• A. A line
• B. A circle with centre $z_0$ and radius $r$
• C. A circle with centre $(0,0)$ and radius $1$
• D. A line through origin

Answer 1

The correct option is B A circle with centre $z_0$ and radius $r$

Let $z_0=x_0+i{y}_0$
& $z=x+iy$
$\therefore |z-z_0|=r$
$\therefore |x+iy-(x_0+i{y}_0\left)\right|=r$
Squaring both sides,
${|x+iy-(x_0+i{y}_0\left)\right|}^2=r^2$
$(x-x_0{)}^2+(y-y_0{)}^2=r^2$
This is the equation of a circle with radius $r$ and $(x_0,y_0)$ as centre.