What is (x^2-4x)+(y^2+2y)=31 ?

The solution to (x^2-4x)+(y^2+2y)=31 is Circle with (a,b)=(2,-1),r=6

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)=31

(x^{2} - 4 x) + (y^{2} + 2 y) = 31

(a, b) = (2, - 1), r = 6

Solution steps

x^{2} - 4 x + (y^{2} + 2 y) = 31

Rewrite

x^{2} - 4 x + (y^{2} + 2 y) = 31

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 1))^{2} = 6^{2}

Therefore the circle properties are:

(a, b) = (2, - 1), r = 6

(x + 9)^{2} + (y + 2)^{2} = 4

x^{2} + y^{2} + 6 x - 4 y - 36 = 0

(x + 2)^{2} + (y - 3)^{2} = 4^{2}

x^{(2)} + y^{(2)} + 2 x + 4 y - 15 = 0

(x - 3)^{2} + (y - 5)^{2} = 306.25

What is (x^2-4x)+(y^2+2y)=31 ?
The solution to (x^2-4x)+(y^2+2y)=31 is Circle with (a,b)=(2,-1),r=6