What is x^2+y^2-x+8y-3=0 ?

The solution to x^2+y^2-x+8y-3=0 is Circle with (a,b)=(1/2 ,-4),r=(sqrt(77))/2

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x^2+y^2-x+8y-3=0

x^{2} + y^{2} - x + 8 y - 3 = 0

(a, b) = (\frac{1}{2}, - 4), r = \frac{77}{2}

Solution steps

x^{2} + y^{2} - x + 8 y - 3 = 0

Rewrite

x^{2} + y^{2} - x + 8 y - 3 = 0

in the form of the standard circle equation

(x - \frac{1}{2})^{2} + (y - (- 4))^{2} = (\frac{77}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, - 4), r = \frac{77}{2}

4 x^{(2)} + 4 y^{(2)} + 6 x + 8 y - 5 = 0

(x + 2)^{2} + (y - 6)^{2} = 10

x^{2} + y^{2} - 6 x + 8 y + 20 = 0

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

2 x^{2} + 2 y^{2} + 8 x - 20 y - 18 = 0

What is x^2+y^2-x+8y-3=0 ?
The solution to x^2+y^2-x+8y-3=0 is Circle with (a,b)=(1/2 ,-4),r=(sqrt(77))/2