SOLUTION: How do you find the center, radius, and x and y intercepts of 3x^2-3y^2-6x-12y=0?

Algebra ->  Graphs -> SOLUTION: How do you find the center, radius, and x and y intercepts of 3x^2-3y^2-6x-12y=0?      Log On


   

Question 334204: How do you find the center, radius, and x and y intercepts of 3x^2-3y^2-6x-12y=0?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Complete the square in x and y.
3x%5E2-3y%5E2-6x-12y=0
x%5E2-2x-y%5E2-4y=0
x%5E2-2x%2B1-1=y%5E2%2B4y%2B4-4
%28x-1%29%5E2-1=%28y%2B2%29%5E2-4
%28y%2B2%29%5E2-%28x-1%29%5E2=5
%28y%2B2%29%5E2%2F5-%28x-1%29%5E2%2F5=1
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It's not a circle, it's a hyperbola, so it has no radius.
It's centered at (1,2).
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x%5E2-2x-y%5E2-4y=0
To find the x-intercepts, set y=0 and solve for x.
x%5E2-2x-y%5E2-4y=0
x%5E2-2x=0
x%28x-2%29=0
(0,0) and (2,0)
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To find the y-intercepts, set x=0 and solve for y.
x%5E2-2x-y%5E2-4y=0
-y%5E2-4y=0
y%28y%2B4%29=0
(0,0) and (0,-4)
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