What is (x+8/3)^2+y^2=1 ?

The solution to (x+8/3)^2+y^2=1 is Circle with (a,b)=(-8/3 ,0),r=1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x+8/3)^2+y^2=1

(x + \frac{8}{3})^{2} + y^{2} = 1

(a, b) = (- \frac{8}{3}, 0), r = 1

Solution steps

(x + \frac{8}{3})^{2} + y^{2} = 1

Rewrite

(x + \frac{8}{3})^{2} + y^{2} = 1

in the form of the standard circle equation

(x - (- \frac{8}{3}))^{2} + (y - 0)^{2} = 1^{2}

Therefore the circle properties are:

(a, b) = (- \frac{8}{3}, 0), r = 1

x^{2} + y^{2} - 6 x - 8 y - 24 = 0

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

y^{2} - 4 y + x^{2} - 6 x - 12 = 0

2 x^{2} + 2 y^{2} - 10 x - 18 y = 1

a^{2} + b^{2} = a + b

What is (x+8/3)^2+y^2=1 ?
The solution to (x+8/3)^2+y^2=1 is Circle with (a,b)=(-8/3 ,0),r=1