What is (x+1)^2+(y+3)^2=11 ?

The solution to (x+1)^2+(y+3)^2=11 is Circle with (a,b)=(-1,-3),r=sqrt(11)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x+1)^2+(y+3)^2=11

(x + 1)^{2} + (y + 3)^{2} = 11

(a, b) = (- 1, - 3), r = 11

Solution steps

(x + 1)^{2} + (y + 3)^{2} = 11

Rewrite

(x + 1)^{2} + (y + 3)^{2} = 11

in the form of the standard circle equation

(x - (- 1))^{2} + (y - (- 3))^{2} = (11)^{2}

Therefore the circle properties are:

(a, b) = (- 1, - 3), r = 11

x^{2} + y^{2} + 10 y + 11 = 0

x^{2} + y^{2} + 10 x - 8 y - 6 = 0

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

x^{2} + y^{2} - 8 x - 4 y + 4 = 0

4 x^{2} + 4 y^{2} - 16 x + 20 y + 25 = 0

What is (x+1)^2+(y+3)^2=11 ?
The solution to (x+1)^2+(y+3)^2=11 is Circle with (a,b)=(-1,-3),r=sqrt(11)