What is the center 4x^2+8x+4y^2-20y=35 ?

The center 4x^2+8x+4y^2-20y=35 is (-1, 5/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center 4x^2+8x+4y^2-20y=35

center

4 x^{2} + 8 x + 4 y^{2} - 20 y = 35

(- 1, \frac{5}{2})

Solution steps

4 x^{2} + 8 x + 4 y^{2} - 20 y = 35

Rewrite

4 x^{2} + 8 x + 4 y^{2} - 20 y = 35

in the form of the standard circle equation

(x - (- 1))^{2} + (y - \frac{5}{2})^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (- 1, \frac{5}{2}), r = 4

And the center is:

(- 1, \frac{5}{2})

3 x^{2} + 3 y^{2} - 18 x - 30 y - 6 = 0

(x - 4)^{2} + (y - 1)^{2} = 218

4 x^{2} + 4 y^{2} - x + 2 y - 11 = 0

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

X^{2} + Y^{2} + 40 X - 10 Y + 55 = 0

What is the center 4x^2+8x+4y^2-20y=35 ?
The center 4x^2+8x+4y^2-20y=35 is (-1, 5/2)