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

Popular >

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)

3 x^{2} + 3 y^{2} = 75

x^{2} + y^{2} - 5 x + 4 y - 46 = 0

3 (x + 1)^{2} + 3 (y - 1)^{2} = 6

x^{2} + y^{2} - 8 x + 12 = 0

x^{2} + y^{2} - 6 x - 8 y - 11 = 0

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