How do you find the center and radius of the circle #6x^2 + 6y^2 + 12x − y = 0#?

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Answer 1 · 2017-01-12T20:41:01

#C(-1;1/12)#

and #r=sqrt(145)/12#

You would write the given equation in the standard form:

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

Then it is, by dividing all terms by 6,

#x^2+y^2+2x-1/6y=0#

You would find the center and radius by using the following formulas:

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

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

Then #C(-2/2;-(-1/6)/2)=(-1;1/12)#

and #r=1/2sqrt(2^2+(-1/6)^2-4*0)=1/2sqrt(4+1/36)=1/2sqrt(145/36)=sqrt(145)/12#