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

The radius x^2+y^2-2x+4y-4=0 is r=3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

radius x^2+y^2-2x+4y-4=0

radius

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

r = 3

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 1)^{2} + (y - (- 2))^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (1, - 2), r = 3

And the radius is:

r = 3

x^{2} + y^{2} = 3

x^{2} + y^{2} + 14 x - 2 y + 25 = 0

x^{2} + y^{2} - 12 x + 20 y + 15 = 0

16 x^{2} + 16 y^{2} + 8 x - 32 y = - 1

x^{2} + y^{2} - 2 x + 6 y - 15 = 0