What is (x^2+2x)+(y^2+4y)=-1 ?

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

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(x^2+2x)+(y^2+4y)=-1

(x^{2} + 2 x) + (y^{2} + 4 y) = - 1

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

Solution steps

x^{2} + 2 x + (y^{2} + 4 y) = - 1

Rewrite

x^{2} + 2 x + (y^{2} + 4 y) = - 1

in the form of the standard circle equation

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

Therefore the circle properties are:

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

x^{2} + y^{2} - 4 x - 6 y - 36 = 0

(x + 9)^{2} + (y - 8)^{2} = 81

x, x^{2} + Y (x)^{2} = 100 + 5 Y

2 x^{2} + 2 y^{2} + 12 x - 8 y + 14 = 0

x^{2} + y^{2} + 6 x - 12 y + 40 = 0

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