Find Center and Radius by General Equation of Circle (METHOD 1)

How to find Center and Radius of a Circle by its General Equation (METHOD 1)

Click here for METHOD 2 

EXAMPLE: Find center and radius of the circle with equation

x² + y² – 8x + 2y + 8 = 0

SOLUTION:

METHOD 1: 

The General Equation of a Circle is

x² + y² + 2gx + 2f y + c = 0 ————— (1)

where, 

(–g, -f ) is the center of a circle and

 `sqrt{g^2+f^2-c}`  is the radius of a circle. 

Step 1: 

x² + y² – 8x + 2y + 8 = 0        (given)

Rewrite the terms of given equation of circle

x² + y² + 2(- 4) x + 2(1)y + 8 = 0 ———-(2)

Step 2: Compare equation (1) and equation (2)

we get

g = -4

f = 1

c = 8

Step 3: 

Thus, a circle whose equation is

x² + y² – 8x + 2y + 8 = 0

has center at (-g, –f ) = (4, -1)

and has radius `sqrt{g^2+f^2-c}`  =  `sqrt{(-4)^2+1^2-8}`

= `sqrt{16+1-8}`

=`sqrt{9}`

= 3

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