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

The solution to x^2+y^2-4x+2y=0 is Circle with (a,b)=(2,-1),r=sqrt(5)

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=0

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

(a, b) = (2, - 1), r = 5

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 1))^{2} = (5)^{2}

Therefore the circle properties are:

(a, b) = (2, - 1), r = 5

x^{2} + y^{2} - \frac{20}{17} x - \frac{72}{17} y - \frac{912}{17} = 0

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

X^{2} + Y^{2} + 40 X - 10 Y + 55 = 0

x^{2} + y^{2} + 6 x - 10 y - 17 = 0

(x + 5)^{2} + (y + 7)^{2} = 2 1^{2}

What is x^2+y^2-4x+2y=0 ?
The solution to x^2+y^2-4x+2y=0 is Circle with (a,b)=(2,-1),r=sqrt(5)