What is (x^2+4x)+(y^2+4y)=9 ?

The solution to (x^2+4x)+(y^2+4y)=9 is Circle with (a,b)=(-2,-2),r=sqrt(17)

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

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

(a, b) = (- 2, - 2), r = 17

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 2))^{2} + (y - (- 2))^{2} = (17)^{2}

Therefore the circle properties are:

(a, b) = (- 2, - 2), r = 17

x^{2} + y^{2} + 12 x + 2 y + \frac{1}{16} = 0

(x - 3)^{2} + (y + 2)^{2} = 17

x^{2} + y^{2} - 12 x = - 4 y + 5

(3 x - 2 - 2)^{(2)} + (3 y + 1)^{(2)} = 9

x^{2} + y^{2} - 8 x + 6 y = - 24

What is (x^2+4x)+(y^2+4y)=9 ?
The solution to (x^2+4x)+(y^2+4y)=9 is Circle with (a,b)=(-2,-2),r=sqrt(17)