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

The center x^2+y^2-12x+4y=-2 is (6,-2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center x^2+y^2-12x+4y=-2

center

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

(6, - 2)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 6)^{2} + (y - (- 2))^{2} = (38)^{2}

Therefore the circle properties are:

(a, b) = (6, - 2), r = 38

And the center is:

(6, - 2)

x^{2} + y^{2} + 10 x - 6 y + 15 = 0

a^{2} + b^{2} = 36

(x + 5)^{2} + (y - 3)^{2} = 25

9 x^{2} + 9 y^{2} + 42 x - 54 y + 49 = 0

(x - (- 5))^{2} + (y - (7))^{2} = (11)^{2}