What is x^2-8x+y^2-2y-19=0 ?

The solution to x^2-8x+y^2-2y-19=0 is Circle with (a,b)=(4,1),r=6

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x^2-8x+y^2-2y-19=0

x^{2} - 8 x + y^{2} - 2 y - 19 = 0

(a, b) = (4, 1), r = 6

Solution steps

x^{2} - 8 x + y^{2} - 2 y - 19 = 0

Rewrite

x^{2} - 8 x + y^{2} - 2 y - 19 = 0

in the form of the standard circle equation

(x - 4)^{2} + (y - 1)^{2} = 6^{2}

Therefore the circle properties are:

(a, b) = (4, 1), r = 6

(x + 5)^{2} + (y + 0.6666600000000 \dots)^{2} = 0.4444400000000 \dots

x^{2} + y^{2} = 19.4 x + 3 y - 4

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

(x - 6)^{2} + (y - 4)^{2} = 16

x^{2} + y^{2} + \frac{2}{3} x - \frac{1}{2} y = 0

What is x^2-8x+y^2-2y-19=0 ?
The solution to x^2-8x+y^2-2y-19=0 is Circle with (a,b)=(4,1),r=6