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

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

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

x^{2} + y^{2} = (22)^{2}

(a, b) = (0, 0), r = (22)^{2}

Solution steps

x^{2} + y^{2} = (22)^{2}

Rewrite

x^{2} + y^{2} = (22)^{2}

in the form of the standard circle equation

(x - 0)^{2} + (y - 0)^{2} = ((22)^{2})^{2}

Therefore the circle properties are:

(a, b) = (0, 0), r = (22)^{2}

(x - 1)^{2} + (y - 4)^{2} = 81

(x + \frac{1}{7})^{2} + (y - \frac{3}{5})^{2} = \frac{25}{9}

(x - 4)^{2} + (y + 12)^{2} = 1 7^{2}

9 x^{2} + 9 y^{2} - 12 x + 36 y = 104

4 x^{2} + 40 x + 4 y^{2} - 32 = 5

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