What is the center 4x^2+4y^2+12x-4y-90=0 ?

The center 4x^2+4y^2+12x-4y-90=0 is (-3/2 , 1/2)

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+12x-4y-90=0

center

4 x^{2} + 4 y^{2} + 12 x - 4 y - 90 = 0

(- \frac{3}{2}, \frac{1}{2})

Solution steps

4 x^{2} + 4 y^{2} + 12 x - 4 y - 90 = 0

Rewrite

4 x^{2} + 4 y^{2} + 12 x - 4 y - 90 = 0

in the form of the standard circle equation

(x - (- \frac{3}{2}))^{2} + (y - \frac{1}{2})^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (- \frac{3}{2}, \frac{1}{2}), r = 5

And the center is:

(- \frac{3}{2}, \frac{1}{2})

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

x^{2} + y^{2} + 4 x - 6 y = 12

x^{2} + (y - 4)^{2} = 25

x^{2} + y^{2} = 3

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

What is the center 4x^2+4y^2+12x-4y-90=0 ?
The center 4x^2+4y^2+12x-4y-90=0 is (-3/2 , 1/2)