What is the center 4x^2+16x+15+4y^2+6y=0 ?

The center 4x^2+16x+15+4y^2+6y=0 is (-2,-3/4)

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center 4x^2+16x+15+4y^2+6y=0

center

4 x^{2} + 16 x + 15 + 4 y^{2} + 6 y = 0

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

Solution steps

4 x^{2} + 16 x + 15 + 4 y^{2} + 6 y = 0

Rewrite

4 x^{2} + 16 x + 15 + 4 y^{2} + 6 y = 0

in the form of the standard circle equation

(x - (- 2))^{2} + (y - (- \frac{3}{4}))^{2} = (\frac{13}{4})^{2}

Therefore the circle properties are:

(a, b) = (- 2, - \frac{3}{4}), r = \frac{13}{4}

And the center is:

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

(x + 4)^{2} + (y + 3)^{2} = 25

(x - 1)^{2} + (y - 3)^{2} = 9

5 x^{2} + 3 x + 5 y^{2} - 2 y - 9 = 0

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

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

What is the center 4x^2+16x+15+4y^2+6y=0 ?
The center 4x^2+16x+15+4y^2+6y=0 is (-2,-3/4)