What is x^2+y^2+2y-48=0 ?

The solution to x^2+y^2+2y-48=0 is Circle with (a,b)=(0,-1),r=7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

x^2+y^2+2y-48=0

x^{2} + y^{2} + 2 y - 48 = 0

(a, b) = (0, - 1), r = 7

Solution steps

x^{2} + y^{2} + 2 y - 48 = 0

Rewrite

x^{2} + y^{2} + 2 y - 48 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 1))^{2} = 7^{2}

Therefore the circle properties are:

(a, b) = (0, - 1), r = 7

What is x^2+y^2+2y-48=0 ?
The solution to x^2+y^2+2y-48=0 is Circle with (a,b)=(0,-1),r=7