What is 9-(x-5)^2=(y^2+4y+4) ?

The solution to 9-(x-5)^2=(y^2+4y+4) is Circle with (a,b)=(5,-2),r=3

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9-(x-5)^2=(y^2+4y+4)

9 - (x - 5)^{2} = (y^{2} + 4 y + 4)

(a, b) = (5, - 2), r = 3

Solution steps

9 - (x - 5)^{2} = (y^{2} + 4 y + 4)

Rewrite

9 - (x - 5)^{2} = (y^{2} + 4 y + 4)

in the form of the standard circle equation

(x - 5)^{2} + (y - (- 2))^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (5, - 2), r = 3

x^{2} + 4 x + y^{2} - 2 y = 11

x^{2} + y^{2} + 2 x - 10 y - 104 = 0

x^{2} + y^{2} - 6 x + 4 y = 3

(x + 1)^{2} + (y + 4)^{2} = 144

(x - 5)^{2} + (y + 7)^{2} = 100

What is 9-(x-5)^2=(y^2+4y+4) ?
The solution to 9-(x-5)^2=(y^2+4y+4) is Circle with (a,b)=(5,-2),r=3