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

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

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

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

(a, b) = (2, 4), r = 3

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (2, 4), r = 3

(x - 5.5)^{2} + (y - 5.5)^{2} - (1.5)^{2} = 0

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

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

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

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

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