What is 2x^2+2y^2-4x+12y-18=0 ?

The solution to 2x^2+2y^2-4x+12y-18=0 is Circle with (a,b)=(1,-3),r=sqrt(19)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2x^2+2y^2-4x+12y-18=0

2 x^{2} + 2 y^{2} - 4 x + 12 y - 18 = 0

(a, b) = (1, - 3), r = 19

Solution steps

2 x^{2} + 2 y^{2} - 4 x + 12 y - 18 = 0

Rewrite

2 x^{2} + 2 y^{2} - 4 x + 12 y - 18 = 0

in the form of the standard circle equation

(x - 1)^{2} + (y - (- 3))^{2} = (19)^{2}

Therefore the circle properties are:

(a, b) = (1, - 3), r = 19

4 x^{2} + 4 y^{2} + 40 x - 32 y = 5

x^{2} + y^{2} + 8 y = 33

x^{2} + 2 x + y^{2} - 35 = 0

x^{2} + y^{2} + 12 x - 8 y + 3 = 0

8 x^{2} + 8 y^{2} - 20 x + 64 y - 371 = 0

What is 2x^2+2y^2-4x+12y-18=0 ?
The solution to 2x^2+2y^2-4x+12y-18=0 is Circle with (a,b)=(1,-3),r=sqrt(19)