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

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Answer 1 · 2016-01-11T20:45:09

centre = (2 , 1 ) and radius = 3

comparing the general form for the equation of a circle :

#x^2 + y^2 +2gx + 2fy + 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 ) #

#rArr r = sqrt((- 2 )^2 + (- 1 )^2 -( -4 ) ) =sqrt(4 + 1 + 4 )= sqrt9 = 3 #

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