What is (X-3)^2+(Y+1)^2=sqrt(6) ?

The solution to (X-3)^2+(Y+1)^2=sqrt(6) is Circle with (a,b)=(3,-1),r=sqrt(\sqrt{6)}

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(X-3)^2+(Y+1)^2=sqrt(6)

(X - 3)^{2} + (Y + 1)^{2} = 6

(a, b) = (3, - 1), r = 6

Solution steps

(X - 3)^{2} + (Y + 1)^{2} = 6

Rewrite

(X - 3)^{2} + (Y + 1)^{2} = 6

in the form of the standard circle equation

(X - 3)^{2} + (Y - (- 1))^{2} = (6)^{2}

Therefore the circle properties are:

(a, b) = (3, - 1), r = 6

x^{2} + 4 x = 10 - y^{2}

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

x^{2} + y^{2} = 9

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

x^{2} + y^{2} = 49

What is (X-3)^2+(Y+1)^2=sqrt(6) ?
The solution to (X-3)^2+(Y+1)^2=sqrt(6) is Circle with (a,b)=(3,-1),r=sqrt(\sqrt{6)}