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# Can you solve this circle by completing the square?

3x^2+3y^2-6x+5y=0 Find the center, and radius

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- 1 decade agoFavorite Answer
a circle has the following equation:

(x-x0)^2 + (y-y0)^2 = r^2

where (x0, y0) are the coordinates of the center and r is the radius

you have:

3x^2+3y^2-6x+5y=0

first divide by three

x^2+y^2-2x+5/3y=0

than you must have two perfect squares

x^2-2x+1-1 +y^2+5/3y +(5/6)^2-(5/6)^2=0

(x-1)^2 + (y+5/6)^2 -1-(5/6)^2=0

(x-1)^2 + (y-(-5/6))^2 = 1+25/36 = 61/36

(x-1)^2 + (y-(-5/6))^2 = 61/36

the circle has the center in (1,-5/6) and a radius sqrt(61)/6

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