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

The solution to 4x^2+4y^2+8y-3=0 is Circle with (a,b)=(0,-1),r=(sqrt(7))/2

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

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

(a, b) = (0, - 1), r = \frac{7}{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 1))^{2} = (\frac{7}{2})^{2}

Therefore the circle properties are:

(a, b) = (0, - 1), r = \frac{7}{2}

(x + 5)^{2} + (y - 1)^{2} = 1

(x + 3)^{2} + (y + 1)^{2} = \frac{4}{25}

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

x^{2} + y^{2} - 8 x + 12 y + 36 = 0

x^{2} + y^{2} + 8 x + 2 y + 16 = 0

What is 4x^2+4y^2+8y-3=0 ?
The solution to 4x^2+4y^2+8y-3=0 is Circle with (a,b)=(0,-1),r=(sqrt(7))/2