What is x^2+y^2+12x+10y+41=0 ?

The solution to x^2+y^2+12x+10y+41=0 is Circle with (a,b)=(-6,-5),r=2sqrt(5)

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x^2+y^2+12x+10y+41=0

x^{2} + y^{2} + 12 x + 10 y + 41 = 0

(a, b) = (- 6, - 5), r = 25

Solution steps

x^{2} + y^{2} + 12 x + 10 y + 41 = 0

Rewrite

x^{2} + y^{2} + 12 x + 10 y + 41 = 0

in the form of the standard circle equation

(x - (- 6))^{2} + (y - (- 5))^{2} = (25)^{2}

Therefore the circle properties are:

(a, b) = (- 6, - 5), r = 25

(x - 4)^{2} + (y + 1)^{2} = 5

(x + 3)^{2} + (y - 7)^{2} = 9

(x - 3)^{2} + 2 y + 1 - 81 = - y^{2}

(x + 4)^{2} + (y + 2)^{2} = 16

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

What is x^2+y^2+12x+10y+41=0 ?
The solution to x^2+y^2+12x+10y+41=0 is Circle with (a,b)=(-6,-5),r=2sqrt(5)