What is x^2+y^2+2/3 x-1/2 y=0 ?

The solution to x^2+y^2+2/3 x-1/2 y=0 is Circle with (a,b)=(-1/3 , 1/4),r= 5/12

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2+y^2+2/3 x-1/2 y=0

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

(a, b) = (- \frac{1}{3}, \frac{1}{4}), r = \frac{5}{12}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- \frac{1}{3}))^{2} + (y - \frac{1}{4})^{2} = (\frac{5}{12})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{3}, \frac{1}{4}), r = \frac{5}{12}

x^{2} + y^{2} + 20 x + 20 y + 119 = 0

(x - 3)^{2} + (y - 4)^{2} = 49

x^{2} + y^{2} - 10 x - 8 y - 23 = 0

1 \leq x^{2} + y^{2}

(x - 8)^{2} + (y + 2)^{2} = 324

What is x^2+y^2+2/3 x-1/2 y=0 ?
The solution to x^2+y^2+2/3 x-1/2 y=0 is Circle with (a,b)=(-1/3 , 1/4),r= 5/12