What is the radius x^2-2x+y^2-8y+1=0 ?

The radius x^2-2x+y^2-8y+1=0 is r=4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

radius x^2-2x+y^2-8y+1=0

radius

x^{2} - 2 x + y^{2} - 8 y + 1 = 0

r = 4

Solution steps

x^{2} - 2 x + y^{2} - 8 y + 1 = 0

Rewrite

x^{2} - 2 x + y^{2} - 8 y + 1 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (1, 4), r = 4

And the radius is:

r = 4

x^{2} + y^{2} - 8 x + 12 y + 10 = 0

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

(x - \frac{3}{2})^{2} + (y - \frac{3}{2})^{2} = \frac{9}{4}

x^{2} + y^{2} - 8 x + 12 = 0

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