What is x^2+(y+5)^2=27 ?

The solution to x^2+(y+5)^2=27 is Circle with (a,b)=(0,-5),r=sqrt(27)

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x^2+(y+5)^2=27

x^{2} + (y + 5)^{2} = 27

(a, b) = (0, - 5), r = 27

Solution steps

x^{2} + (y + 5)^{2} = 27

Rewrite

x^{2} + (y + 5)^{2} = 27

in the form of the standard circle equation

(x - 0)^{2} + (y - (- 5))^{2} = (27)^{2}

Therefore the circle properties are:

(a, b) = (0, - 5), r = 27

3 x^{2} + 3 y^{2} + 6 x + 9 y - 12 = 0

x^{2} + y^{2} - 4 y - 6 y + 9 = 0

3 x^{2} = - 3 y^{2} + 12

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

x^{2} + y^{2} - 2 \geq 0

What is x^2+(y+5)^2=27 ?
The solution to x^2+(y+5)^2=27 is Circle with (a,b)=(0,-5),r=sqrt(27)