What is (x-3)^2+(y-1)^2=2 ?

The solution to (x-3)^2+(y-1)^2=2 is Circle with (a,b)=(3,1),r=sqrt(2)

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(x-3)^2+(y-1)^2=2

(x - 3)^{2} + (y - 1)^{2} = 2

(a, b) = (3, 1), r = 2

Solution steps

(x - 3)^{2} + (y - 1)^{2} = 2

Rewrite

(x - 3)^{2} + (y - 1)^{2} = 2

in the form of the standard circle equation

(x - 3)^{2} + (y - 1)^{2} = (2)^{2}

Therefore the circle properties are:

(a, b) = (3, 1), r = 2

(x^{2} - 4 x) + (y^{2} - 6 y) = 23

(x + 2)^{(2)} + (y + 5)^{(2)} = 144

x^{2} + 4 x + 2^{2} + y^{2} + 8 y + 4^{2} - 5 = 20

x^{2} + y^{2} - 5 x - 3 y = 0

9 x^{2} + 72 x + 9 y^{2} - 12 y + 103 = 0

What is (x-3)^2+(y-1)^2=2 ?
The solution to (x-3)^2+(y-1)^2=2 is Circle with (a,b)=(3,1),r=sqrt(2)