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

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

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

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

(a, b) = (0, 0), r = 2

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, 0), r = 2

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

(x + 3)^{2} + (y - 5)^{2} = 4^{2}

x^{2} + 5 x + y^{2} - 7 y + \frac{247}{16} = 0

(x - 5)^{(2)} + (y + 7)^{(2)} = 50

x^{2} - 16 x + 64 + y^{2} - 16 x = 0

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