What is 10(x+5)^2+10(y-8)^2=1000 ?

The solution to 10(x+5)^2+10(y-8)^2=1000 is Circle with (a,b)=(-5,8),r=10

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

10(x+5)^2+10(y-8)^2=1000

10 (x + 5)^{2} + 10 (y - 8)^{2} = 1000

(a, b) = (- 5, 8), r = 10

Solution steps

10 (x + 5)^{2} + 10 (y - 8)^{2} = 1000

Rewrite

10 (x + 5)^{2} + 10 (y - 8)^{2} = 1000

in the form of the standard circle equation

(x - (- 5))^{2} + (y - 8)^{2} = 1 0^{2}

Therefore the circle properties are:

(a, b) = (- 5, 8), r = 10

x^{2} + y^{2} + 2 x + 12 y + 36 = 0

(x - 2)^{2} + (y + 9)^{2} = 3 9^{2}

x^{2} + (y + 8)^{2} = 400

x^{2} + y^{2} - 2 x - 10 y + 17 = 0

x^{2} + y^{2} - 2 x + 10 y + 19 = 0

What is 10(x+5)^2+10(y-8)^2=1000 ?
The solution to 10(x+5)^2+10(y-8)^2=1000 is Circle with (a,b)=(-5,8),r=10