What is the center 4x^2+4y^2-16x-24y+51=0 ?

The center 4x^2+4y^2-16x-24y+51=0 is (2,3)

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center 4x^2+4y^2-16x-24y+51=0

center

4 x^{2} + 4 y^{2} - 16 x - 24 y + 51 = 0

(2, 3)

Solution steps

4 x^{2} + 4 y^{2} - 16 x - 24 y + 51 = 0

Rewrite

4 x^{2} + 4 y^{2} - 16 x - 24 y + 51 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (2, 3), r = \frac{1}{2}

And the center is:

(2, 3)

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

2 x^{2} + 2 y^{2} + 10 x - 6 y - 55 = 0

x^{2} + y^{2} - 12 x - 10 y = - 12

x^{2} + y^{2} = 5

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

What is the center 4x^2+4y^2-16x-24y+51=0 ?
The center 4x^2+4y^2-16x-24y+51=0 is (2,3)