What is 4x^2+4y^2+8x-4y-3=0 ?

The solution to 4x^2+4y^2+8x-4y-3=0 is Circle with (a,b)=(-1, 1/2),r=sqrt(2)

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4x^2+4y^2+8x-4y-3=0

4 x^{2} + 4 y^{2} + 8 x - 4 y - 3 = 0

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

Solution steps

4 x^{2} + 4 y^{2} + 8 x - 4 y - 3 = 0

Rewrite

4 x^{2} + 4 y^{2} + 8 x - 4 y - 3 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

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

x^{2} + y^{2} - 25 x = - 1 - 3 y

(x - 11)^{2} + (y - 8)^{2} = 268

4 x^{2} - 24 x + 4 y^{2} + 20 y = 39

x^{2} + (y + 1)^{2} = 64

What is 4x^2+4y^2+8x-4y-3=0 ?
The solution to 4x^2+4y^2+8x-4y-3=0 is Circle with (a,b)=(-1, 1/2),r=sqrt(2)