What is the center 9x^2+36x+9y^2+72y+155=0 ?

The center 9x^2+36x+9y^2+72y+155=0 is (-2,-4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center 9x^2+36x+9y^2+72y+155=0

center

9 x^{2} + 36 x + 9 y^{2} + 72 y + 155 = 0

(- 2, - 4)

Solution steps

9 x^{2} + 36 x + 9 y^{2} + 72 y + 155 = 0

Rewrite

9 x^{2} + 36 x + 9 y^{2} + 72 y + 155 = 0

in the form of the standard circle equation

(x - (- 2))^{2} + (y - (- 4))^{2} = (\frac{5}{3})^{2}

Therefore the circle properties are:

(a, b) = (- 2, - 4), r = \frac{5}{3}

And the center is:

(- 2, - 4)

(x - 2)^{2} + (y - 5)^{2} = 49

\frac{( x + 7 ) ^{2}}{16} + \frac{( y + 5 ) ^{2}}{16} = 1

x^{2} + 4 x + y^{2} + 6 y + 9 = 0

(x - 2)^{2} + (y + 6)^{2} = 25

(x - 1)^{2} + y^{2} = 4

What is the center 9x^2+36x+9y^2+72y+155=0 ?
The center 9x^2+36x+9y^2+72y+155=0 is (-2,-4)