What is the center 2x^2+2y^2+16x+20y+18=0 ?

The center 2x^2+2y^2+16x+20y+18=0 is (-4,-5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center 2x^2+2y^2+16x+20y+18=0

center

2 x^{2} + 2 y^{2} + 16 x + 20 y + 18 = 0

(- 4, - 5)

Solution steps

2 x^{2} + 2 y^{2} + 16 x + 20 y + 18 = 0

Rewrite

2 x^{2} + 2 y^{2} + 16 x + 20 y + 18 = 0

in the form of the standard circle equation

(x - (- 4))^{2} + (y - (- 5))^{2} = (42)^{2}

Therefore the circle properties are:

(a, b) = (- 4, - 5), r = 42

And the center is:

(- 4, - 5)

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

x^{2} + y^{2} + 10 x + 12 y + 25 = 0

(x + 12)^{2} + (y - 4)^{2} - 30 = 0

x^{2} + y^{2} - 18 x - 6 y - 13 = 8

(x + 1)^{2} + (y - 2)^{2} = 25

What is the center 2x^2+2y^2+16x+20y+18=0 ?
The center 2x^2+2y^2+16x+20y+18=0 is (-4,-5)