Question 1207943: A. Find the center and radius of the circle.
B. Graph the circle.
C. Find the intercepts of the circle, if any.
3x^2 + 3y^2 - 12y = 0
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! A. Find the center and radius of the circle.
B. Graph the circle.
C. Find the intercepts of the circle, if any.
3x^2 + 3y^2 - 12y = 0
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x^2 + y^2 - 4y = 0
x^2 + y^2 - 4y + 4 = 4
(x-0)^2 + (y-2)^2 = 2^2
Center is (0,2)
r = 2
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You can graph it on paper, or
DL the FREE graph software from several sites.
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x^2 + y^2 - 4y = 0
Sub zero for x to find the y-intercepts
y^2 - 4y = 0
y*(y-4) = 0
y = 0 --> (0,0)
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y = 4 --> (0,4)
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Sub zero for y to find the x-intercepts
x^2 + y^2 - 4y = 0
x^2 = 0
--> (0,0)
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