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

The solution to x^2-x+y^2+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-x+y^2+y=0

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

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

Solution steps

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

Rewrite

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

\frac{y ^{2}}{3.5 8 ^{2}} + \frac{x ^{2}}{3.5 8 ^{2}} = 1

8^{2} = a^{2} + b^{2}

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

(x - 1)^{2} + (y - (- 3))^{2} = 6. 7^{2}

2 x^{2} + 2 y^{2} + 12 x - 28 y + 104 = 0

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