What is the center x^2+y^2+8x+4y-3=40 ?

The center x^2+y^2+8x+4y-3=40 is (-4,-2)

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center x^2+y^2+8x+4y-3=40

center

x^{2} + y^{2} + 8 x + 4 y - 3 = 40

(- 4, - 2)

Solution steps

x^{2} + y^{2} + 8 x + 4 y - 3 = 40

Rewrite

x^{2} + y^{2} + 8 x + 4 y - 3 = 40

in the form of the standard circle equation

(x - (- 4))^{2} + (y - (- 2))^{2} = (37)^{2}

Therefore the circle properties are:

(a, b) = (- 4, - 2), r = 37

And the center is:

(- 4, - 2)

(x - 1)^{2} = - 4 (y - 2)

\frac{x ^{2}}{16} - \frac{y ^{2}}{4} = 1

x^{2} + \frac{y ^{2}}{49} = 1

\frac{x ^{2}}{4} - \frac{y ^{2}}{25} = 1

y^{2} = 16 x

What is the center x^2+y^2+8x+4y-3=40 ?
The center x^2+y^2+8x+4y-3=40 is (-4,-2)