What is (x-4)^2+(y-4)^2=3^2 ?

The solution to (x-4)^2+(y-4)^2=3^2 is Circle with (a,b)=(4,4),r=3

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(x-4)^2+(y-4)^2=3^2

(x - 4)^{2} + (y - 4)^{2} = 3^{2}

(a, b) = (4, 4), r = 3

Solution steps

(x - 4)^{2} + (y - 4)^{2} = 3^{2}

Rewrite

(x - 4)^{2} + (y - 4)^{2} = 3^{2}

in the form of the standard circle equation

(x - 4)^{2} + (y - 4)^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (4, 4), r = 3

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

2 x^{2} + 2 y^{2} + 14 x - 10 y + 35 = 0

x^{2} + y^{2} + x + 6 y - 55 = 0

x^{2} + y^{2} - 8 x - 6 y - 19 = 0

(x - \frac{23}{5})^{2} + (y + \frac{2}{5})^{2} = \frac{373}{25}

What is (x-4)^2+(y-4)^2=3^2 ?
The solution to (x-4)^2+(y-4)^2=3^2 is Circle with (a,b)=(4,4),r=3