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

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

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x^2+y^2-4=0

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

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

Solution steps

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

Rewrite

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

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 = - \frac{1}{2} (y + 3)^{2} + 1

\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1

4 x^{2} + 9 y^{2} = 36

3 x^{2} + 6 y^{2} = 18

x^{2} = - 12 y

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