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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

center

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

(1, 2)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(1, 2)

x^{2} + y^{2} + 6 x - 8 y + 16 = 0

9 x^{2} + 9 y^{2} = 729

x^{2} + y^{2} + 3 x + 4 y + 4 = 0

(x - 9)^{2} + (y + 4)^{2} = 25

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