What is the center 3x^2+3y^2-27x-9y-24=0 ?

The center 3x^2+3y^2-27x-9y-24=0 is (9/2 , 3/2)

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center 3x^2+3y^2-27x-9y-24=0

center

3 x^{2} + 3 y^{2} - 27 x - 9 y - 24 = 0

(\frac{9}{2}, \frac{3}{2})

Solution steps

3 x^{2} + 3 y^{2} - 27 x - 9 y - 24 = 0

Rewrite

3 x^{2} + 3 y^{2} - 27 x - 9 y - 24 = 0

in the form of the standard circle equation

(x - \frac{9}{2})^{2} + (y - \frac{3}{2})^{2} = (\frac{61}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{9}{2}, \frac{3}{2}), r = \frac{61}{2}

And the center is:

(\frac{9}{2}, \frac{3}{2})

2 x^{2} + 2 y^{2} - 16 x + 4 y - 16 = 0

(x - 1)^{2} + (y + 3)^{2} = 16

9 x^{2} + 12 x + 9 y^{2} - 18 y - 23 = 0

x^{2} + y^{2} - 8 x - 12 y + 43 = 0

(x - 1)^{2} + (y + 3)^{2} = 25

What is the center 3x^2+3y^2-27x-9y-24=0 ?
The center 3x^2+3y^2-27x-9y-24=0 is (9/2 , 3/2)