What is x^2+y^2+2x-10y-5=0 ?

The solution to x^2+y^2+2x-10y-5=0 is Circle with (a,b)=(-1,5),r=sqrt(31)

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x^2+y^2+2x-10y-5=0

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

(a, b) = (- 1, 5), r = 31

Solution steps

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

Rewrite

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

in the form of the standard circle equation

(x - (- 1))^{2} + (y - 5)^{2} = (31)^{2}

Therefore the circle properties are:

(a, b) = (- 1, 5), r = 31

x^{2} + 4 x + y^{2} - y = \frac{3}{5}

x^{(2)} + y^{(2)} + 4 x - 2 y + 1 = 0

(x^{2} - 4 x) + (y^{2} + 8 y) = 16

x^{2} + y^{2} - 6 x + 10 y + 27 = 0

x^{2} + y^{2} - 12 x - 12 y + 36 = 0

What is x^2+y^2+2x-10y-5=0 ?
The solution to x^2+y^2+2x-10y-5=0 is Circle with (a,b)=(-1,5),r=sqrt(31)