What is the radius x^2+y^2-4x+8y-5=0 ?

The radius x^2+y^2-4x+8y-5=0 is r=5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

radius x^2+y^2-4x+8y-5=0

radius

x^{2} + y^{2} - 4 x + 8 y - 5 = 0

r = 5

Solution steps

x^{2} + y^{2} - 4 x + 8 y - 5 = 0

Rewrite

x^{2} + y^{2} - 4 x + 8 y - 5 = 0

in the form of the standard circle equation

(x - 2)^{2} + (y - (- 4))^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (2, - 4), r = 5

And the radius is:

r = 5

x^{2} + y^{2} + 8 x - 9 y = 6

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

x^{2} + y^{2} + 6 x - 4 y - 12 = 0

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

x^{2} + y^{2} - 8 x - 6 y - 11 = 0