What is x^2+y^2-x-y=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

(a, b) = (\frac{1}{2}, \frac{1}{2}), r = \frac{1}{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, \frac{1}{2}), r = \frac{1}{2}

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

x^{2} + y^{2} - 8 x - 10 y + 40 = 0

x^{2} + y^{2} - 18 x - 14 y + 66 = 0

x^{2} + y^{2} - 4 x + 8 y + 11 = 0

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

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