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

The solution to x^2+y^2-2x<= 8 is Circle with (a,b)=(1,0),r=3

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

x^{2} + y^{2} - 2 x \leq 8

(a, b) = (1, 0), r = 3

Solution steps

x^{2} + y^{2} - 2 x \leq 8

Rewrite

x^{2} + y^{2} - 2 x \leq 8

in the form of the standard circle equation

(x - 1)^{2} + (y - 0)^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (1, 0), r = 3

(x + 2)^{2} + (y + 4)^{2} = 9

(x - 1)^{(2)} + (y - 5)^{(2)} = 16

4 y^{2} + 4 x^{2} - 24 x + 32 y = - 84

x^{2} + y^{2} + 14 x - 7 y + 40 = 0

9 x^{2} + 36 x + 9 y^{2} + 72 y = - 155

What is x^2+y^2-2x<= 8 ?
The solution to x^2+y^2-2x<= 8 is Circle with (a,b)=(1,0),r=3