What is the center 4x^2+4y^2+36y+5=0 ?

The center 4x^2+4y^2+36y+5=0 is (0,-9/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center 4x^2+4y^2+36y+5=0

center

4 x^{2} + 4 y^{2} + 36 y + 5 = 0

(0, - \frac{9}{2})

Solution steps

4 x^{2} + 4 y^{2} + 36 y + 5 = 0

Rewrite

4 x^{2} + 4 y^{2} + 36 y + 5 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - (- \frac{9}{2}))^{2} = (19)^{2}

Therefore the circle properties are:

(a, b) = (0, - \frac{9}{2}), r = 19

And the center is:

(0, - \frac{9}{2})

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

x^{2} + y^{2} - 18 x + 8 y + 81 = 0

(x - 2)^{2} + y^{2} = 16

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

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