Asked by Kim George
find the center and radii of the circle 3x^2-3y^2+4x-5y+2=0?
Answers
Answered by
oobleck
In the first place, x^2-y^2+... is not a circle. But, given that you know it's an hyperbola, just complete the squares to find its standard form.
3x^2-3y^2+4x-5y+2=0
3x^2+4x - 3y^2-5y = -2
x^2 + 4/3 x - y^2 - 5/3 y = -2/3
(x^2 + 4/3 x + (2/3)^2) - (y^2 + 5/3 y + (5/6)^2) = -2/3 + (2/3)^2 - (5/6)^2
(x + 2/3)^2 - (y + 5/6)^2 = -11/12
(y + 5/6)^2 / (11/12) - (x + 2/3)^2 / (11/12) = 1
3x^2-3y^2+4x-5y+2=0
3x^2+4x - 3y^2-5y = -2
x^2 + 4/3 x - y^2 - 5/3 y = -2/3
(x^2 + 4/3 x + (2/3)^2) - (y^2 + 5/3 y + (5/6)^2) = -2/3 + (2/3)^2 - (5/6)^2
(x + 2/3)^2 - (y + 5/6)^2 = -11/12
(y + 5/6)^2 / (11/12) - (x + 2/3)^2 / (11/12) = 1
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