What is x^2+y^2+1x-4y-8=0 ?

The solution to x^2+y^2+1x-4y-8=0 is Circle with (a,b)=(-1/2 ,2),r= 7/2

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x^2+y^2+1x-4y-8=0

x^{2} + y^{2} + 1 x - 4 y - 8 = 0

(a, b) = (- \frac{1}{2}, 2), r = \frac{7}{2}

Solution steps

x^{2} + y^{2} + 1 \cdot x - 4 y - 8 = 0

Rewrite

x^{2} + y^{2} + 1 x - 4 y - 8 = 0

in the form of the standard circle equation

(x - (- \frac{1}{2}))^{2} + (y - 2)^{2} = (\frac{7}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{1}{2}, 2), r = \frac{7}{2}

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

(X - 3)^{2} + (Y + 1)^{2} = 6

x^{2} + 4 x = 10 - y^{2}

x^{2} + y^{2} = 6 x + 6 y

x^{2} + y^{2} = 9

What is x^2+y^2+1x-4y-8=0 ?
The solution to x^2+y^2+1x-4y-8=0 is Circle with (a,b)=(-1/2 ,2),r= 7/2