What is 4x^2+4y^2-64=0 ?

The solution to 4x^2+4y^2-64=0 is Circle with (a,b)=(0,0),r=4

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4x^2+4y^2-64=0

4 x^{2} + 4 y^{2} - 64 = 0

(a, b) = (0, 0), r = 4

Solution steps

4 x^{2} + 4 y^{2} - 64 = 0

Rewrite

4 x^{2} + 4 y^{2} - 64 = 0

in the form of the standard circle equation

(x - 0)^{2} + (y - 0)^{2} = 4^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = 4

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

x^{2} - 8 x + 16 + y^{2} - 18 y + 81 = 48

(x + 3)^{2} + (y - 2)^{2} = 49

x^{2} - 8 x + 16 + y^{2} - 18 y + 81 = 16

x^{2} + y^{2} - 6 x - 2 y + 9 = 0

What is 4x^2+4y^2-64=0 ?
The solution to 4x^2+4y^2-64=0 is Circle with (a,b)=(0,0),r=4