What is 5x^2-5x+5y^2-4y-9=0 ?

The solution to 5x^2-5x+5y^2-4y-9=0 is Circle with (a,b)=(1/2 , 2/5),r=(sqrt(221))/(10)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5x^2-5x+5y^2-4y-9=0

5 x^{2} - 5 x + 5 y^{2} - 4 y - 9 = 0

(a, b) = (\frac{1}{2}, \frac{2}{5}), r = \frac{221}{10}

Solution steps

5 x^{2} - 5 x + 5 y^{2} - 4 y - 9 = 0

Rewrite

5 x^{2} - 5 x + 5 y^{2} - 4 y - 9 = 0

in the form of the standard circle equation

(x - \frac{1}{2})^{2} + (y - \frac{2}{5})^{2} = (\frac{221}{10})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, \frac{2}{5}), r = \frac{221}{10}

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

x^{2} + y^{2} \leq 4 x + 4 y

3 x^{2} + 3 y^{2} = y + 2

x^{2} + (y - 115)^{2} = 11025

y^{2} + x^{2} + 24 x + 10 y + 160 = 0

What is 5x^2-5x+5y^2-4y-9=0 ?
The solution to 5x^2-5x+5y^2-4y-9=0 is Circle with (a,b)=(1/2 , 2/5),r=(sqrt(221))/(10)