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 65137: Find the center, radius, and x- and y-intercepts of the circle given by the equation 3x2 + 3y2 - 6x + 5y = 0.
Answer by stanbon(75887) (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.
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Divide thru by 3 to get:
x^2-2x+y^2+(5/3)y=0
Complete the square, as follows:
(x^2-2x+1)+(y^2+(5/3)y+(5/6)^2)=1+(25/36)
(x-1)^2+(y+(5/6))^2 =61/36
Center: (1,-(5/6))
Radius: [sqrt(61)]/6
x-intercepts:
Let y=0 then x^2-2x=0; then x=0 or x=2; Intercepts at (0,0) and (2,0)
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y-intercepts:
Let x=0, then y^2+(5/3)y=0; y=0 or y=(-5/3): Interceps at (0,0), and (0,-5/3)
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Cheers,
Stan H.
3(x^2-2x+1)+3(y^2+(5/3)y+(5/6)^2) = 3+3(5/6)^2
3(x-1)^2 + 3(y+(5/6))^2 = 108/36 + 25/36
3(x-1)^2 + 3(y+(5/6))^2 = 133/36
(x-1)^2 + (y+(5/6))^2= 133/108

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