What is the center 4x^2+4y^2-16x+24y-12=0 ?

The center 4x^2+4y^2-16x+24y-12=0 is (2,-3)

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center 4x^2+4y^2-16x+24y-12=0

center

4 x^{2} + 4 y^{2} - 16 x + 24 y - 12 = 0

(2, - 3)

Solution steps

4 x^{2} + 4 y^{2} - 16 x + 24 y - 12 = 0

Rewrite

4 x^{2} + 4 y^{2} - 16 x + 24 y - 12 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 3))^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (2, - 3), r = 4

And the center is:

(2, - 3)

x^{2} + 10 x + y^{2} - 16 y - 11 = 0

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

x^{2} + y^{2} + 8 x - 12 y + 27 = 0

x^{2} + y^{2} + 8 y = 33

x^{2} + y^{2} - 4 x + 2 y - 11 = 0

What is the center 4x^2+4y^2-16x+24y-12=0 ?
The center 4x^2+4y^2-16x+24y-12=0 is (2,-3)