What is 4x^2+4y^2-16y+15=0 ?

The solution to 4x^2+4y^2-16y+15=0 is Circle with (a,b)=(0,2),r= 1/2

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4x^2+4y^2-16y+15=0

4 x^{2} + 4 y^{2} - 16 y + 15 = 0

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

Solution steps

4 x^{2} + 4 y^{2} - 16 y + 15 = 0

Rewrite

4 x^{2} + 4 y^{2} - 16 y + 15 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

2 x^{2} + 8 x + 2 y^{2} - 18 y = - \frac{33}{2}

x^{2} + 4 x + y^{2} - 14 y = - 3

(x - 1)^{2} + (y + 2)^{2} + (x - 4)^{2} + (y - 6)^{2} = 73

x^{2} + y^{2} + 28 x - 32 y + 448 = 0

3 x^{2} + 3 y^{2} = 243

What is 4x^2+4y^2-16y+15=0 ?
The solution to 4x^2+4y^2-16y+15=0 is Circle with (a,b)=(0,2),r= 1/2