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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

center

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

(2, 1)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

And the center is:

(2, 1)

(x - 1)^{2} + (y + 2)^{2} = 16

x^{2} + y^{2} = 8 (x + y) + 4

(\frac{x - 7}{4})^{2} + (\frac{y + 3}{4})^{2} = \frac{169}{16}

x^{2} + y^{2} + 4 x + 6 y + 9 = 0

x^{2} - 14 x + 49 + y^{2} + 8 y + 16 - 25 = 0