What is the center 4x^2+4y^2+40x-32y=5 ?

The center 4x^2+4y^2+40x-32y=5 is (-5,4)

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+40x-32y=5

center

4 x^{2} + 4 y^{2} + 40 x - 32 y = 5

(- 5, 4)

Solution steps

4 x^{2} + 4 y^{2} + 40 x - 32 y = 5

Rewrite

4 x^{2} + 4 y^{2} + 40 x - 32 y = 5

in the form of the standard circle equation

(x - (- 5))^{2} + (y - 4)^{2} = (\frac{13}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 5, 4), r = \frac{13}{2}

And the center is:

(- 5, 4)

4 x^{2} + 4 y^{2} + 8 x - y = 0

(x + 2)^{2} + (y + 5)^{2} = 16

x^{2} + (y - 3)^{2} = 16

4 x^{2} + 4 y^{2} - 16 = 0

x^{2} + y^{2} + 12 y + 32 = 0

What is the center 4x^2+4y^2+40x-32y=5 ?
The center 4x^2+4y^2+40x-32y=5 is (-5,4)