What is 3x^2+3y^2+6x-3y-2=0 ?

The solution to 3x^2+3y^2+6x-3y-2=0 is Circle with (a,b)=(-1, 1/2),r=(sqrt(23))/(2sqrt(3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3x^2+3y^2+6x-3y-2=0

3 x^{2} + 3 y^{2} + 6 x - 3 y - 2 = 0

(a, b) = (- 1, \frac{1}{2}), r = \frac{23}{2 3}

Solution steps

3 x^{2} + 3 y^{2} + 6 x - 3 y - 2 = 0

Rewrite

3 x^{2} + 3 y^{2} + 6 x - 3 y - 2 = 0

in the form of the standard circle equation

(x - (- 1))^{2} + (y - \frac{1}{2})^{2} = (\frac{23}{2 3})^{2}

Therefore the circle properties are:

(a, b) = (- 1, \frac{1}{2}), r = \frac{23}{2 3}

y^{2} = - x^{2} + 25

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

a^{2} + b^{2} = 2

x^{2} + y^{2} - 10 x - 4 y + 20 = 0

(x + 6)^{2} + (y - 7)^{2} = 36

What is 3x^2+3y^2+6x-3y-2=0 ?
The solution to 3x^2+3y^2+6x-3y-2=0 is Circle with (a,b)=(-1, 1/2),r=(sqrt(23))/(2sqrt(3))