How do you find the center and radius of the circle #9x^2+9y^2+6y-107=0#?

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Answer 1 · 2016-06-17T21:43:39

#C(0;-1/3)#; #r=2sqrt(3)#

you can write the equivalent equation by dividing all terms by 9:

#x^2+y^2+2/3y-107/9=0#

you obtain the center of the circle by using the formulas:

#C(-a/2;-b/2)#

and radius by

#r=1/2sqrt(a^2+b^2-4c)#

where a, b, c are in the following general circle equation:

#x^2+y^2+ax+by+c=0#

So
#C(0;-2/3*1/2)#

#C(0;-1/3)#

#r=1/2sqrt(4/9-4(-107/9))#

#r=1/2sqrt(4/9+428/9)#

#r=1/2sqrt(432/9)#

#r=1/2sqrt(48)#

#r=1/2*4sqrt(3)#

#r=2sqrt(3)#