What is (x-4)^2+y^2-4=0 ?

The solution to (x-4)^2+y^2-4=0 is Circle with (a,b)=(4,0),r=2

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(x-4)^2+y^2-4=0

(x - 4)^{2} + y^{2} - 4 = 0

(a, b) = (4, 0), r = 2

Solution steps

(x - 4)^{2} + y^{2} - 4 = 0

Rewrite

(x - 4)^{2} + y^{2} - 4 = 0

in the form of the standard circle equation

(x - 4)^{2} + (y - 0)^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (4, 0), r = 2

x^{2} + y^{2} + 4 x - 6 y + 3 = 0

(x - 13.4)^{2} + (y + 2.6)^{2} = 100

(x - 3)^{2} + (y + 5)^{2} = 36

x^{2} + y^{2} + 18 x + 4 y - 15 = 0

x^{2} + y^{2} + 2 x + 14 y - 31 = 0

What is (x-4)^2+y^2-4=0 ?
The solution to (x-4)^2+y^2-4=0 is Circle with (a,b)=(4,0),r=2