3x^2+3y^2-6x+18y+18=0

what is the standard form of this equation of a circle? 3x^2+3y^2-6x+18y+18=0

 

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1 Answer

3x^2 -6x +3y^2 +18y =-18

3(x^2 -2x + ___ ) +3(y^2 +6y + ___ ) =-18 +3(__ ) +3( __ )

3(x^2-2x+1) +3(y^2 +6y +9) =-18 +3(1) +3(9)

3(x-1)^2 +3(y+3)^2 =12

(x-1)^2 +(y+3)^2 =4
by Level 1 User (940 points)

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