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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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