What is x^2+(y-2)^2=sqrt(28) ?

The solution to x^2+(y-2)^2=sqrt(28) is Circle with (a,b)=(0,2),r=sqrt(\sqrt{28)}

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x^2+(y-2)^2=sqrt(28)

x^{2} + (y - 2)^{2} = 28

(a, b) = (0, 2), r = 28

Solution steps

x^{2} + (y - 2)^{2} = 28

Rewrite

x^{2} + (y - 2)^{2} = 28

in the form of the standard circle equation

(x - 0)^{2} + (y - 2)^{2} = (28)^{2}

Therefore the circle properties are:

(a, b) = (0, 2), r = 28

x^{2} + y^{2} - 8 x + 6 y + 21 = 0

1 - x^{2} - y^{2} \geq 0

(x - 6)^{2} + (y - 4)^{2} - 64 = 0

x^{2} + y^{2} + 8 x - 6 y + 16 = 0

(x - 7)^{2} + (y - 3)^{2} = 7

What is x^2+(y-2)^2=sqrt(28) ?
The solution to x^2+(y-2)^2=sqrt(28) is Circle with (a,b)=(0,2),r=sqrt(\sqrt{28)}