What is x^2+4x+y^2-2y=11 ?

The solution to x^2+4x+y^2-2y=11 is Circle with (a,b)=(-2,1),r=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2+4x+y^2-2y=11

x^{2} + 4 x + y^{2} - 2 y = 11

(a, b) = (- 2, 1), r = 4

Solution steps

x^{2} + 4 x + y^{2} - 2 y = 11

Rewrite

x^{2} + 4 x + y^{2} - 2 y = 11

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 2, 1), r = 4

x^{2} + y^{2} + 2 x - 10 y - 104 = 0

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

(x + 1)^{2} + (y + 4)^{2} = 144

(x - 5)^{2} + (y + 7)^{2} = 100

x^{2} + y^{2} + 12 x + 6 y + 41 = 0

What is x^2+4x+y^2-2y=11 ?
The solution to x^2+4x+y^2-2y=11 is Circle with (a,b)=(-2,1),r=4