What is 5x^2+5y^2-4x+y-3=0 ?

The solution to 5x^2+5y^2-4x+y-3=0 is Circle with (a,b)=(2/5 ,-1/10),r=(sqrt(77))/(10)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5x^2+5y^2-4x+y-3=0

5 x^{2} + 5 y^{2} - 4 x + y - 3 = 0

(a, b) = (\frac{2}{5}, - \frac{1}{10}), r = \frac{77}{10}

Solution steps

5 x^{2} + 5 y^{2} - 4 x + y - 3 = 0

Rewrite

5 x^{2} + 5 y^{2} - 4 x + y - 3 = 0

in the form of the standard circle equation

(x - \frac{2}{5})^{2} + (y - (- \frac{1}{10}))^{2} = (\frac{77}{10})^{2}

Therefore the circle properties are:

(a, b) = (\frac{2}{5}, - \frac{1}{10}), r = \frac{77}{10}

x^{2} + y^{2} + 6 x - 8 y - 10 = 0

3^{2} = (x - 0)^{2} + (y + 2)^{2}

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

9 = x^{2} - 6 x + 9 + y^{2} + 4 y + 4

4 x^{2} + 4 y^{2} = 256

What is 5x^2+5y^2-4x+y-3=0 ?
The solution to 5x^2+5y^2-4x+y-3=0 is Circle with (a,b)=(2/5 ,-1/10),r=(sqrt(77))/(10)