What is 4x^2+4y^2+4x-4y+1=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4x^2+4y^2+4x-4y+1=0

4 x^{2} + 4 y^{2} + 4 x - 4 y + 1 = 0

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

Solution steps

4 x^{2} + 4 y^{2} + 4 x - 4 y + 1 = 0

Rewrite

4 x^{2} + 4 y^{2} + 4 x - 4 y + 1 = 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}

2 x^{2} + 2 y^{2} + 16 - 40 = 0

(x + 5)^{2} + (y + 7)^{2} = 4

10 x^{2} + 10 y^{2} - 200 = 0

(x + 6)^{2} + (y + 14)^{2} = 100

3 x^{2} + 3 y^{2} + 12 x - 22 y + 35 = 0

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