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.
How to Find 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.
|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
|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
|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.
|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.
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