What is x^2-8x+y^2=0 ?

The solution to x^2-8x+y^2=0 is Circle with (a,b)=(4,0),r=4

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x^2-8x+y^2=0

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} - 8 y + 12 = 0

(x - 6)^{2} + (y - 10)^{2} = (5)^{2}

(x + 6)^{2} + (y + 8)^{2} = 100

3 x^{2} + 3 y^{2} - 12 x - 24 y + 12 = 0

(x - 5)^{2} + (y - 7)^{2} = 4

What is x^2-8x+y^2=0 ?
The solution to x^2-8x+y^2=0 is Circle with (a,b)=(4,0),r=4