# HOW TO FIND CENTER AND RADIUS FROM AN EQUATION IN COMPLEX NUMBERS

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.

## Equation of the Circle from Complex Numbers

The locus of z that satisfies the equation |z − z0| = r where z0 is a fixed complex number and r is a fixed positive real number consists of all points z whose distance from z0 is r . So, |z − z0| = r is the complex form of the equation of a circle.

(i) |z − z0| < r represents the points interior of the circle.

(ii) |z − z0> r represents the points exterior of the circle.

Note :

|z| = r ==> √x2 + y2  =  r

x2 + y2  =  r2, represents a circle centre at the origin with radius r units.

## Finding Centre and Radius of Circle From Complex Numbers  - Examples

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 − z0| = 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 − z0| = 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 − z0| = 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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