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

The solution to x^2+y^2+10x-10y+41=0 is Circle with (a,b)=(-5,5),r=3

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

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

(a, b) = (- 5, 5), r = 3

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (- 5, 5), r = 3

5 x^{2} + x + 5 y^{2} - y - 9 = 0

x^{2} - 4 x + y^{2} - 6 y = 3

(x - 9)^{2} + (y - 2)^{2} = 25

(x + 9)^{2} + (y - 3)^{2} = 80

x^{2} + y^{2} + 4 x - 20 y + 40 = 0

What is x^2+y^2+10x-10y+41=0 ?
The solution to x^2+y^2+10x-10y+41=0 is Circle with (a,b)=(-5,5),r=3