What is 2x^2+2y^2-8x+12y-1=0 ?

The solution to 2x^2+2y^2-8x+12y-1=0 is Circle with (a,b)=(2,-3),r=(3sqrt(3))/(sqrt(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2x^2+2y^2-8x+12y-1=0

2 x^{2} + 2 y^{2} - 8 x + 12 y - 1 = 0

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

Solution steps

2 x^{2} + 2 y^{2} - 8 x + 12 y - 1 = 0

Rewrite

2 x^{2} + 2 y^{2} - 8 x + 12 y - 1 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

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

x^{2} + y^{2} = 40

x^{2} + y^{2} = 21

x^{2} + y^{2} = 30

x^{2} + y^{2} + 6 x + 10 y - 87 = 0

What is 2x^2+2y^2-8x+12y-1=0 ?
The solution to 2x^2+2y^2-8x+12y-1=0 is Circle with (a,b)=(2,-3),r=(3sqrt(3))/(sqrt(2))