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

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

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

(a, b) = (\frac{1}{2}, 0), r = \frac{1}{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, 0), r = \frac{1}{2}

x^{2} + y^{2} - 6 x + 10 y + 25 = 0

(x - 1)^{2} + (y - 7)^{2} = 25

(x - 15)^{2} + ((y - (- 20)))^{2} = 9^{2}

(x - 12)^{2} + (y - 2)^{2} = 64

x^{2} + y^{2} - 14 x + 2 y + 10 = 0

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