What is the center x^2+8x+y^2+2y+8=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center x^2+8x+y^2+2y+8=0

center

x^{2} + 8 x + y^{2} + 2 y + 8 = 0

(- 4, - 1)

Solution steps

x^{2} + 8 x + y^{2} + 2 y + 8 = 0

Rewrite

x^{2} + 8 x + y^{2} + 2 y + 8 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 4, - 1), r = 3

And the center is:

(- 4, - 1)

x^{2} + 6 x + y^{2} - 3 y = - \frac{1}{4}

x^{2} + y^{2} - 3 x + 4 y = 0

x^{2} + 4 x + 10 y + y^{2} + 3 = 0

2 x^{2} + 2 y^{2} - 8 = 0

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