Question

The equation $\mid z - {z_o}| = r$ represents (Given: ${z_o}$ is a fixed complex number.)
A. A line
B. A circle with centre ${z_o}$ and radius r
C. A circle with centre (0, 0) and radius 1
D. A line through origin

Answer 1

Start by taking $z = x + iy$ and ${z_o} = {x_0} + i{y_o}$ as $z$ is a changing or locus which we have to find and ${z_o}$ being a fixed complex number. Take this and find the modulus.

Complete answer:
We are given that this equation $\mid z - {z_o}| = r$ . We have to find the nature of $z$ .
First we need to understand what a fixed point is and what a moving point is.

Here, we are given that ${z_o}$ is a fixed complex number. This means that this complex number represents only a single point on the complex plane.

Whereas, a moving point is such which represents a curve on the complex plane as given in this question as $z$ . This $z$ is bounded by certain conditions by which it makes a curve here this condition is $\mid z - {z_o}| = r$ .

Now, let us start by assuming –
${z_o} = {x_0} + i{y_o}$ (Where${x_0}$and${y_0}$are constants)
$z = x + iy$ (Where$x$and$y$are variables)

Substituting the values in the given condition we have,
$\mid z - {z_o}| = r$
$|x + iy - ({x_0} + i{y_o})| = r$

Subtracting real from real and imaginary from imaginary we get,
$|(x - {x_0}) + i(y - {y_o})| = r$
Taking the modulus of the complex number we get,
$\sqrt {{{(x - {x_0})}^2} + {{(y - {y_o})}^2}} = r$

Squaring both the sides we get,
${(x - {x_0})^2} + {(y - {y_o})^2} = {r^2}$
This is the equation of a circle with its centre at $({x_o},{y_o})$ and the radius equal to r.

Since, the point $({x_o},{y_o})$ is represented by the complex number ${z_o}$
So, we can say that the centre of the circle is ${z_o}$ .

Hence, the correct answer is option (B).

Note: We can solve it in a smaller way. We know the $z$ is a locus of points and ${z_o}$ is a fixed point. The condition $\mid z - {z_o}| = r$ is the distance between the $z$ and ${z_o}$ which is always equal to r which is true for only the type of curve which is a circle with its centre at ${z_o}$ .