What is x^2+(y+1)^2=2^2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2+(y+1)^2=2^2

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

(a, b) = (0, - 1), r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, - 1), r = 2

5 0^{2} = (x - 21.21)^{2} + (y - 21.21)^{2}

x^{2} + y^{2} - 4 x - 14 y + 50 = 0

9 x^{2} + 9 y^{2} + 72 y + 36 y + 155 = 0

(x + 1)^{2} + (y - 2)^{2} = 36

(x + 7)^{2} + (y + 4)^{2} = 3

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