What is 4x^2+4y^2+8y-45=0 ?

The solution to 4x^2+4y^2+8y-45=0 is Circle with (a,b)=(0,-1),r= 7/2

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4x^2+4y^2+8y-45=0

4 x^{2} + 4 y^{2} + 8 y - 45 = 0

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

Solution steps

4 x^{2} + 4 y^{2} + 8 y - 45 = 0

Rewrite

4 x^{2} + 4 y^{2} + 8 y - 45 = 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, h (x)^{2} + (8 - x)^{2} = 49

a^{2} + b^{2} = 7^{2}

x^{2} + y^{2} - 18 x + 12 y = - 108

(- 7 - p)^{2} + (0 - q)^{2} = 625

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

What is 4x^2+4y^2+8y-45=0 ?
The solution to 4x^2+4y^2+8y-45=0 is Circle with (a,b)=(0,-1),r= 7/2