**How to Find Center and Radius From an Equation in Complex Numbers :**

Here we are going to see some example problems based on finding center and radius from an equation in complex numbers.

The locus of z that satisfies the equation |z − z_{0}| = r where z_{0} is a fixed complex number and r is a fixed positive real number consists of all points z whose distance from z_{0} is r .

So, |z − z_{0}| = r is the complex form of the equation of a circle.

(i) |z − z_{0}| < r represents the points interior of the circle.

(ii) |z − z_{0}| > r represents the points exterior of the circle.

Note :

|z| = r ==> √x^{2} + y^{2} = r

x^{2} + y^{2} = r^{2}, represents a circle centre at the origin with radius r units.

**Question 1 :**

Show that the following equations represent a circle, and, find its centre and radius

(i) |z - 2 - i| = 3

**Solution :**

|z - 2 - i| = 3

|z - (2 + i)| = 3

It is of the form |z − z_{0}| = r and so it represents a circle, whose centre and radius are (2, 1) and 3 respectively.

(ii) |2z + 2 − 4i| = 2

**Solution :**

|2z + 2 − 4i| = 2

2|z + (2 − 4i)/2| = 2

|z + (2 − 4i)/2| = 1

|z - (-(1 - 2i))| = 1

|z - (-1 + 2i)| = 1

It is of the form |z − z_{0}| = r and so it represents a circle, whose centre and radius are (-1, 2) and 1 respectively.

(iii) |3z − 6 +12i| = 8.

**Solution :**

|3z − 6 +12i| = 8

3|z - (6 − 12i)/3| = 8

|z - (2 − 4i)| = 8/3

It is of the form |z − z_{0}| = r and so it represents a circle, whose centre and radius are (2, -4) and 8/3 respectively.

After having gone through the stuff given above, we hope that the students would have understood, "How to Find Center and Radius From an Equation in Complex Numbers".

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