What is x^2+y^2+8x+8y+4=0 ?

The solution to x^2+y^2+8x+8y+4=0 is Circle with (a,b)=(-4,-4),r=2sqrt(7)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2+y^2+8x+8y+4=0

x^{2} + y^{2} + 8 x + 8 y + 4 = 0

(a, b) = (- 4, - 4), r = 27

Solution steps

x^{2} + y^{2} + 8 x + 8 y + 4 = 0

Rewrite

x^{2} + y^{2} + 8 x + 8 y + 4 = 0

in the form of the standard circle equation

(x - (- 4))^{2} + (y - (- 4))^{2} = (27)^{2}

Therefore the circle properties are:

(a, b) = (- 4, - 4), r = 27

4 x^{2} + 4 y^{2} + 40 x + 16 y + 40 = 0

(x + 1)^{2} + (y + 3)^{2} = 70. 9^{2}

(x - 21)^{2} + (y - 2)^{2} = 16

x^{2} - 2 x + y^{2} - 8 y - 20 = 0

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

What is x^2+y^2+8x+8y+4=0 ?
The solution to x^2+y^2+8x+8y+4=0 is Circle with (a,b)=(-4,-4),r=2sqrt(7)