What is the center x^2+y^2+2x-6y-y=0 ?

The center x^2+y^2+2x-6y-y=0 is (-1, 7/2)

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+2x-6y-y=0

center

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

(- 1, \frac{7}{2})

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 1))^{2} + (y - \frac{7}{2})^{2} = (\frac{53}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 1, \frac{7}{2}), r = \frac{53}{2}

And the center is:

(- 1, \frac{7}{2})

(x + 6)^{2} + (y + 14)^{2} = 400

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

x^{2} + y^{2} + 2 x + 18 y - 143 = 0

x^{2} + y^{2} - 2 x - 4 y + 1 = 0

x^{2} + y^{2} - 2 x - 15 = 0

What is the center x^2+y^2+2x-6y-y=0 ?
The center x^2+y^2+2x-6y-y=0 is (-1, 7/2)