What is the radius x^2+y^2-4x+6y-12=0 ?

The radius x^2+y^2-4x+6y-12=0 is r=5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

radius x^2+y^2-4x+6y-12=0

radius

x^{2} + y^{2} - 4 x + 6 y - 12 = 0

r = 5

Solution steps

x^{2} + y^{2} - 4 x + 6 y - 12 = 0

Rewrite

x^{2} + y^{2} - 4 x + 6 y - 12 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (2, - 3), r = 5

And the radius is:

r = 5

x^{2} + y^{2} + 8 y = 2 x + 20

(x - 2)^{2} + (y - 5)^{2} = 49

x^{2} - 6 x + y^{2} = 7

(x^{2} - 2) + (y^{2} + 4) = 16

4 x^{2} + 4 x + 4 y^{2} - 16 y - 19 = 0