What is x^2+2x+1+y^2+4y+4-36=0 ?

The solution to x^2+2x+1+y^2+4y+4-36=0 is Circle with (a,b)=(-1,-2),r=6

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x^2+2x+1+y^2+4y+4-36=0

x^{2} + 2 x + 1 + y^{2} + 4 y + 4 - 36 = 0

(a, b) = (- 1, - 2), r = 6

Solution steps

x^{2} + 2 x + 1 + y^{2} + 4 y + 4 - 36 = 0

Rewrite

x^{2} + 2 x + 1 + y^{2} + 4 y + 4 - 36 = 0

in the form of the standard circle equation

(x - (- 1))^{2} + (y - (- 2))^{2} = 6^{2}

Therefore the circle properties are:

(a, b) = (- 1, - 2), r = 6

(x - 3)^{2} + (y - 4)^{2} - 9 = y - x + 1

(x - 1)^{2} + (y + 11)^{2} = 100

9 x^{(2)} + 9 y^{(2)} - 6 x + 24 y = 19

x^{2} + y^{2} - 6 x - 6 y = - 5

(x - 0)^{2} + (y - \frac{1}{2})^{2} = (\frac{1}{2})^{2}

What is x^2+2x+1+y^2+4y+4-36=0 ?
The solution to x^2+2x+1+y^2+4y+4-36=0 is Circle with (a,b)=(-1,-2),r=6