What is the center 2x^2+2y^2+12x-2y-3=0 ?

The center 2x^2+2y^2+12x-2y-3=0 is (-3, 1/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

center 2x^2+2y^2+12x-2y-3=0

center

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

Solution steps

2 x^{2} + 2 y^{2} + 12 x - 2 y - 3 = 0

Rewrite

2 x^{2} + 2 y^{2} + 12 x - 2 y - 3 = 0

in the form of the standard circle equation

(x - (- 3))^{2} + (y - \frac{1}{2})^{2} = (\frac{43}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 3, \frac{1}{2}), r = \frac{43}{2}

And the center is:

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

What is the center 2x^2+2y^2+12x-2y-3=0 ?
The center 2x^2+2y^2+12x-2y-3=0 is (-3, 1/2)