What is the center 4x^2+4y^2+20x-8y+9=0 ?

The center 4x^2+4y^2+20x-8y+9=0 is (-5/2 ,1)

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center 4x^2+4y^2+20x-8y+9=0

center

4 x^{2} + 4 y^{2} + 20 x - 8 y + 9 = 0

(- \frac{5}{2}, 1)

Solution steps

4 x^{2} + 4 y^{2} + 20 x - 8 y + 9 = 0

Rewrite

4 x^{2} + 4 y^{2} + 20 x - 8 y + 9 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- \frac{5}{2}, 1), r = 5

And the center is:

(- \frac{5}{2}, 1)

(x - 1)^{2} + (y - 5)^{2} = 17

(x - 2)^{2} + (y + 3)^{2} = \frac{16}{9}

x^{2} + y^{2} - 64 = 0

x^{2} + (y - 2)^{2} = 36

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

What is the center 4x^2+4y^2+20x-8y+9=0 ?
The center 4x^2+4y^2+20x-8y+9=0 is (-5/2 ,1)