What is x^2+y^2+12x+6y+41=0 ?

The solution to x^2+y^2+12x+6y+41=0 is Circle with (a,b)=(-6,-3),r=2

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x^2+y^2+12x+6y+41=0

x^{2} + y^{2} + 12 x + 6 y + 41 = 0

(a, b) = (- 6, - 3), r = 2

Solution steps

x^{2} + y^{2} + 12 x + 6 y + 41 = 0

Rewrite

x^{2} + y^{2} + 12 x + 6 y + 41 = 0

in the form of the standard circle equation

(x - (- 6))^{2} + (y - (- 3))^{2} = 2^{2}

Therefore the circle properties are:

(a, b) = (- 6, - 3), r = 2

x^{2} + y^{2} - 10 x + 12 y - 2 = 0

3 x^{2} + 3 y^{2} - 24 x + 12 y - 15 = 0

5 x^{2} + 5 y^{2} - 20 x + 10 y + 21 = 0

x^{2} + y^{2} - 2 x + y - \frac{11}{4} = 0

x^{2} + y^{2} - 2 x - 1 y - 49 = 0

What is x^2+y^2+12x+6y+41=0 ?
The solution to x^2+y^2+12x+6y+41=0 is Circle with (a,b)=(-6,-3),r=2