What is 4x^2+4y^2+8x-8y+7=0 ?

The solution to 4x^2+4y^2+8x-8y+7=0 is Circle with (a,b)=(-1,1),r= 1/2

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4x^2+4y^2+8x-8y+7=0

4 x^{2} + 4 y^{2} + 8 x - 8 y + 7 = 0

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

Solution steps

4 x^{2} + 4 y^{2} + 8 x - 8 y + 7 = 0

Rewrite

4 x^{2} + 4 y^{2} + 8 x - 8 y + 7 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} + 8 x - 4 y + 19 = 0

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

2 x^{2} + 2 y^{2} + 10 x - 2 y = 7

(x - (- 5))^{2} + (y - 7)^{2} = (11)^{2}

x^{2} + y^{2} + 10 x - 8 y + 36 = 0

What is 4x^2+4y^2+8x-8y+7=0 ?
The solution to 4x^2+4y^2+8x-8y+7=0 is Circle with (a,b)=(-1,1),r= 1/2