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}

3 (x + 1)^{2} + 3 (y - 2)^{2} = 12

x^{2} + y^{2} + 6 x + 12 y + 9 = 0

x^{2} + y^{2} - 16 x + 15 = 0

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

x^{2} - 3 x + 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)