What is (x^2-2x)+(y^2+4y)=-2 ?

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

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(x^2-2x)+(y^2+4y)=-2

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

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

Solution steps

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

Rewrite

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

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^{2} + y^{2} + 2 x - 244 = 0

(x - 3)^{2} + y^{2} < 4

3 x^{2} + 3 y^{2} + 36 x - 24 y + 96 = 0

x^{2} + y^{2} - 20 x - 14 y + 100 = 0

(x - 4)^{2} + (Y (x) + 2)^{2} = 4

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