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

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

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(x-0)^2+(y-2)^2=sqrt(4)

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

(a, b) = (0, 2), r = 4

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (0, 2), r = 4

x^{2} + y^{2} - 16 y + 39 = 0

x^{2} + (y^{2} - 3 y - \frac{9}{4}) = 3 + \frac{9}{4}

x^{2} + y^{2} - 1 0^{2} = 0

\frac{1}{25} (x^{2} + y^{2}) = 1

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

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