What is x^2+y^2=2sqrt(3^2) ?

The solution to x^2+y^2=2sqrt(3^2) is Circle with (a,b)=(0,0),r=sqrt(2\sqrt{3^2)}

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x^2+y^2=2sqrt(3^2)

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

(a, b) = (0, 0), r = 2 3^{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, 0), r = 2 3^{2}

- 2 = x^{2} + y^{2} - 6 x + 5

x^{2} + y^{2} - 4 x - 6 x + 9 = 0

x^{2} + y^{2} + 2 x + 4 y - 12 = 0

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

(x - 6)^{2} + (y - 10)^{2} = 5^{2}

What is x^2+y^2=2sqrt(3^2) ?
The solution to x^2+y^2=2sqrt(3^2) is Circle with (a,b)=(0,0),r=sqrt(2\sqrt{3^2)}