What is x^{(2)}+y^{(2)}+4x-2y+1=0 ?

The solution to x^{(2)}+y^{(2)}+4x-2y+1=0 is Circle with (a,b)=(-2,1),r=2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{(2)}+y^{(2)}+4x-2y+1=0

x^{(2)} + y^{(2)} + 4 x - 2 y + 1 = 0

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

Solution steps

x^{2} + y^{2} + 4 x - 2 y + 1 = 0

Rewrite

x^{2} + y^{2} + 4 x - 2 y + 1 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

(x^{2} - 4 x) + (y^{2} + 8 y) = 16

x^{2} + y^{2} - 6 x + 10 y + 27 = 0

x^{2} + y^{2} - 12 x - 12 y + 36 = 0

(x + 3)^{2} + (y - 2)^{2} = 36

4 x^{2} + 4 y^{2} - 64 = 0

What is x^{(2)}+y^{(2)}+4x-2y+1=0 ?
The solution to x^{(2)}+y^{(2)}+4x-2y+1=0 is Circle with (a,b)=(-2,1),r=2