What is the center (x+5)^2+(y+5)^2=64 ?

The center (x+5)^2+(y+5)^2=64 is (-5,-5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center (x+5)^2+(y+5)^2=64

center

(x + 5)^{2} + (y + 5)^{2} = 64

(- 5, - 5)

Solution steps

(x + 5)^{2} + (y + 5)^{2} = 64

Rewrite

(x + 5)^{2} + (y + 5)^{2} = 64

in the form of the standard circle equation

(x - (- 5))^{2} + (y - (- 5))^{2} = 8^{2}

Therefore the circle properties are:

(a, b) = (- 5, - 5), r = 8

And the center is:

(- 5, - 5)

2 x^{2} + 2 y^{2} - 4 x + 12 y - 18 = 0

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

x^{2} + y^{2} + 6 x - 10 y + 18 = 0

x^{2} + 6 x + y^{2} = 7

x^{2} + y^{2} - 4 x + 6 y - 3 = 0

What is the center (x+5)^2+(y+5)^2=64 ?
The center (x+5)^2+(y+5)^2=64 is (-5,-5)