What is the center (x+12)^2+(y-4)^2-30=0 ?

The center (x+12)^2+(y-4)^2-30=0 is (-12,4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

center (x+12)^2+(y-4)^2-30=0

center

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

(- 12, 4)

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 12, 4), r = 30

And the center is:

(- 12, 4)

x^{2} + y^{2} - 18 x - 6 y - 13 = 8

(x + 1)^{2} + (y - 2)^{2} = 25

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

x^{2} + (y - 1)^{2} = 16

x^{2} + y^{2} + 2 x - 8 y - 19 = 0

What is the center (x+12)^2+(y-4)^2-30=0 ?
The center (x+12)^2+(y-4)^2-30=0 is (-12,4)