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

The center x^2-8x+y^2-2y+8=0 is (4,1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

center x^2-8x+y^2-2y+8=0

center

(4, 1)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 4)^{2} + (y - 1)^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (4, 1), r = 3

And the center is:

(4, 1)