What is the center (x-2)^2+(y+4)^2-16=0 ?

The center (x-2)^2+(y+4)^2-16=0 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=0

center

(x - 2)^{2} + (y + 4)^{2} - 16 = 0

(2, - 4)

Solution steps

(x - 2)^{2} + (y + 4)^{2} - 16 = 0

Rewrite

(x - 2)^{2} + (y + 4)^{2} - 16 = 0

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)

x^{2} + y^{2} + 4 x - 4 y - 28 = 0

x^{2} + y^{2} - 6 x + 8 y - 75 = 0

x^{2} + y^{2} + 13 x + 22 = 0

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

30 = x^{2} - 10 x + y^{2} + 6 y

What is the center (x-2)^2+(y+4)^2-16=0 ?
The center (x-2)^2+(y+4)^2-16=0 is (2,-4)