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

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

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

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

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

Solution steps

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

Rewrite

x^{2} + y^{2} - 2 x + 4 y + 2 = 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 - 15)^{2} + (y + 20)^{2} = 81

2 x^{2} + 6 x + 2 y^{2} - 8 y = \frac{11}{2}

x^{2} + y^{2} - 22 x - 12 y + 150 = 0

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

x^{2} + 8 y + y^{2} + 7 = 0

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