What is the center (x+2)^2+(y+4)^2=16 ?

The center (x+2)^2+(y+4)^2=16 is (-2,-4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

center (x+2)^2+(y+4)^2=16

center

(x + 2)^{2} + (y + 4)^{2} = 16

(- 2, - 4)

Solution steps

(x + 2)^{2} + (y + 4)^{2} = 16

Rewrite

(x + 2)^{2} + (y + 4)^{2} = 16

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(- 2, - 4)

What is the center (x+2)^2+(y+4)^2=16 ?
The center (x+2)^2+(y+4)^2=16 is (-2,-4)