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

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

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

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

(a, b) = (\frac{5}{2}, - \frac{9}{2}), r = \frac{3}{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - \frac{5}{2})^{2} + (y - (- \frac{9}{2}))^{2} = (\frac{3}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{5}{2}, - \frac{9}{2}), r = \frac{3}{2}

x^{2} - 6 + y^{2} = 81

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

(x - \frac{33}{4})^{2} + y^{2} = 25

0 > (x - 1)^{2} + y^{2} - 1

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

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