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

Hint: 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 \[{z_o}\] .