What is x^2+y^2+32x+14y+296=0 ?

The solution to x^2+y^2+32x+14y+296=0 is Circle with (a,b)=(-16,-7),r=3

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x^2+y^2+32x+14y+296=0

x^{2} + y^{2} + 32 x + 14 y + 296 = 0

(a, b) = (- 16, - 7), r = 3

Solution steps

x^{2} + y^{2} + 32 x + 14 y + 296 = 0

Rewrite

x^{2} + y^{2} + 32 x + 14 y + 296 = 0

in the form of the standard circle equation

(x - (- 16))^{2} + (y - (- 7))^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (- 16, - 7), r = 3

(x - 2)^{2} + (y - 3)^{2} = 16

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

x^{2} + y^{2} + 10 x - 6 y + 33 = 0

3 (x - 4)^{2} + 3 y^{2} = 54

x^{2} + y^{2} = 14

What is x^2+y^2+32x+14y+296=0 ?
The solution to x^2+y^2+32x+14y+296=0 is Circle with (a,b)=(-16,-7),r=3