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

The solution to x^2-2x+y^2+4y-4=0 is Circle with (a,b)=(1,-2),r=3

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

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

(a, b) = (1, - 2), r = 3

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (1, - 2), r = 3

(x - 5)^{2} + (y - 3)^{2} = 81

x^{2} + 8 x + y^{2} - 9 y = 6

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

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

4 x^{2} - 8 \geq - 4 y^{2}

What is x^2-2x+y^2+4y-4=0 ?
The solution to x^2-2x+y^2+4y-4=0 is Circle with (a,b)=(1,-2),r=3