What is the center (x-4)^2+(y+1)^2=16 ?

The center (x-4)^2+(y+1)^2=16 is (4,-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

center (x-4)^2+(y+1)^2=16

center

(4, - 1)

Solution steps

(x - 4)^{2} + (y + 1)^{2} = 16

Rewrite

(x - 4)^{2} + (y + 1)^{2} = 16

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(4, - 1)