SOLUTION: Find the center, radius, and x- and y-intercepts of the circle given by the equation 3x2 + 3y2

SOLUTION: Find the center, radius, and x- and y-intercepts of the circle given by the equation 3x2 + 3y2 - 6x + 5y = 0.

Question 64612: Find the center, radius, and x- and y-intercepts of the circle given by the equation 3x2 + 3y2 - 6x + 5y = 0.
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
Find the center, radius, and x- and y-intercepts of the circle given by the equation 3x2 + 3y2 - 6x + 5y = 0.
DIVIDING BY 3
X^2-2X+Y^2+5Y/3=0
(X-1)^2+[Y+(5/6)]^2=1+(5/6)^2=61/36
HENCE CENTER = (1,-5/6)......RADIUS = (1/6)*SQRT(61)
AT Y=0,WE GET
X^2-2X=0=X(X-2)......HENCE X INTERCEPTS ARE 0 AND 2.
AT X=0,WE GET
Y^2+5Y/3=0=Y(Y+5/3)....HENCE Y INTERCEPTS ARE 0 AND -5/3
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