What is (x-3)^2+(y-6)^2=2 ?

The solution to (x-3)^2+(y-6)^2=2 is Circle with (a,b)=(3,6),r=sqrt(2)

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(x-3)^2+(y-6)^2=2

(x - 3)^{2} + (y - 6)^{2} = 2

(a, b) = (3, 6), r = 2

Solution steps

(x - 3)^{2} + (y - 6)^{2} = 2

Rewrite

(x - 3)^{2} + (y - 6)^{2} = 2

in the form of the standard circle equation

(x - 3)^{2} + (y - 6)^{2} = (2)^{2}

Therefore the circle properties are:

(a, b) = (3, 6), r = 2

x^{2} + y^{2} - 10 x + 20 y + 61 = 0

(x - 15)^{2} + (y + 7)^{2} = 225

x^{2} + y^{2} - 2 x - 6 y = 15

\frac{x ^{2} + y ^{2} + 6 x + 3 y + 1}{4} = 0

(x + 9)^{2} + (y - 1)^{2} = 25

What is (x-3)^2+(y-6)^2=2 ?
The solution to (x-3)^2+(y-6)^2=2 is Circle with (a,b)=(3,6),r=sqrt(2)