# How do you find the center and the radius of the circle whose equation is x^2+y^2-4x-2y-4=0?

Jan 11, 2016

centre = (2 , 1 ) and radius = 3

#### Explanation:

comparing the general form for the equation of a circle :

${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$

with the question given

# x^2 + y^2- 4x -2y - 4 = 0

comparing terms we can see that 2g = - 4 hence g = - 2

similarly 2f = - 2 hence f = - 1 and c = - 4

the centre of the circle = (- g , - f ) = (2 , 1 ) and

$r = \sqrt{{g}^{2} + {f}^{2} - c}$

$\Rightarrow r = \sqrt{{\left(- 2\right)}^{2} + {\left(- 1\right)}^{2} - \left(- 4\right)} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$

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