What is the center 4x^2+4y^2+6x+8y-5=0 ?

The center 4x^2+4y^2+6x+8y-5=0 is (-3/4 ,-1)

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center 4x^2+4y^2+6x+8y-5=0

center

4 x^{2} + 4 y^{2} + 6 x + 8 y - 5 = 0

(- \frac{3}{4}, - 1)

Solution steps

4 x^{2} + 4 y^{2} + 6 x + 8 y - 5 = 0

Rewrite

4 x^{2} + 4 y^{2} + 6 x + 8 y - 5 = 0

in the form of the standard circle equation

(x - (- \frac{3}{4}))^{2} + (y - (- 1))^{2} = (\frac{3 5}{4})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{3}{4}, - 1), r = \frac{3 5}{4}

And the center is:

(- \frac{3}{4}, - 1)

x^{2} + y^{2} + 14 x + 16 y + 14 = 0

5 x^{2} + 5 y^{2} + 10 x - 5 y + 3 = 0

(x - 126)^{2} + (y - 102.5)^{2} \geq 149. 5^{2}

(x + 2)^{2} + (y - 5)^{2} = 7

x (x - 12) + y (y + 2) = 12

What is the center 4x^2+4y^2+6x+8y-5=0 ?
The center 4x^2+4y^2+6x+8y-5=0 is (-3/4 ,-1)