find the center and radii of the circle 3x^2-3y^2+4x-5y+2=0?

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Lawrence

1 week ago

1 week ago

the equation is not for a circle, because d coefficient of x^2 and y^2 are not equal. x^2 has 3 while y^2 has -3. so d question is wrong

1 week ago

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