What is (x+2)^2+(y+5)^2=100 ?

The solution to (x+2)^2+(y+5)^2=100 is Circle with (a,b)=(-2,-5),r=10

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x+2)^2+(y+5)^2=100

(x + 2)^{2} + (y + 5)^{2} = 100

(a, b) = (- 2, - 5), r = 10

Solution steps

(x + 2)^{2} + (y + 5)^{2} = 100

Rewrite

(x + 2)^{2} + (y + 5)^{2} = 100

in the form of the standard circle equation

(x - (- 2))^{2} + (y - (- 5))^{2} = 1 0^{2}

Therefore the circle properties are:

(a, b) = (- 2, - 5), r = 10

x^{2} + y^{2} - 16 x + 8 y + 71 = 0

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

9 x^{2} + 9 y^{2} - 6 x - 36 y - 107 = 0

(x - 5)^{2} + (y - 1)^{2} = 25

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

What is (x+2)^2+(y+5)^2=100 ?
The solution to (x+2)^2+(y+5)^2=100 is Circle with (a,b)=(-2,-5),r=10