What is X^2+(y-2)^2=sqrt(2)8 ?

The solution to X^2+(y-2)^2=sqrt(2)8 is Circle with (a,b)=(0,2),r=22^{3/4}

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X^2+(y-2)^2=sqrt(2)8

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

(a, b) = (0, 2), r = 2 2^{\frac{3}{4}}

Solution steps

X^{2} + (y - 2)^{2} = 2 \cdot 8

Rewrite

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

in the form of the standard circle equation

(X - 0)^{2} + (y - 2)^{2} = (2 \cdot 2^{\frac{3}{4}})^{2}

Therefore the circle properties are:

(a, b) = (0, 2), r = 2 2^{\frac{3}{4}}

(x)^{2} + (y + 1)^{2} = 16

x^{2} + 6 x + y^{2} + 2 y - 3 = 0

25 x^{2} + 25 y^{2} + 60 y - 20 x - 360 = 0

(x - 2)^{2} + (y + 8)^{2} = 9

(x + 5)^{2} + (y + 3)^{2} = 100

What is X^2+(y-2)^2=sqrt(2)8 ?
The solution to X^2+(y-2)^2=sqrt(2)8 is Circle with (a,b)=(0,2),r=22^{3/4}