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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

4 y = 5 x - 8

(0, - 2)

(0, 2)

g, T = 2 π \frac{l}{g}

(- 2, 23)

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