What is the center 3x^2+3y^2-24x+6y-24=0 ?

The center 3x^2+3y^2-24x+6y-24=0 is (4,-1)

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center 3x^2+3y^2-24x+6y-24=0

center

3 x^{2} + 3 y^{2} - 24 x + 6 y - 24 = 0

(4, - 1)

Solution steps

3 x^{2} + 3 y^{2} - 24 x + 6 y - 24 = 0

Rewrite

3 x^{2} + 3 y^{2} - 24 x + 6 y - 24 = 0

in the form of the standard circle equation

(x - 4)^{2} + (y - (- 1))^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (4, - 1), r = 5

And the center is:

(4, - 1)

2 x^{2} + 2 y^{2} + 3 x + 5 y - 4 = 0

(x - 1)^{2} + (y - 2)^{2} = 64

x^{2} + (y - 5)^{2} = 25

3 (x - 5)^{2} + 3 y^{2} = 27

x^{2} - 6 x + y^{2} - 8 y = 0

What is the center 3x^2+3y^2-24x+6y-24=0 ?
The center 3x^2+3y^2-24x+6y-24=0 is (4,-1)