What is (x-6)^2+(y-0)^2=3^2 ?

The solution to (x-6)^2+(y-0)^2=3^2 is Circle with (a,b)=(6,0),r=3

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(x-6)^2+(y-0)^2=3^2

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

(a, b) = (6, 0), r = 3

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (6, 0), r = 3

2 x^{2} + 2 y^{2} + 16 x + 12 y + 48 = 0

2 x^{2} + 2 y^{2} - 3 x + y - 1 = 0

3 x^{2} + 3 y^{2} = 3

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

x^{2} + y^{2} + \frac{20}{17} x - \frac{72}{17} y - \frac{912}{17} = 0

What is (x-6)^2+(y-0)^2=3^2 ?
The solution to (x-6)^2+(y-0)^2=3^2 is Circle with (a,b)=(6,0),r=3