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

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

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

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

(a, b) = (\frac{1}{2}, - \frac{1}{2}), r = \frac{3}{2}

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, - \frac{1}{2}), r = \frac{3}{2}

1 (x^{2} + y^{2}) = x

x^{2} + y^{2} + 2 y - 48 = 0

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

2 x^{2} + 2 y^{2} + 8 x - 12 y - 26 = 0

x^{2} + y^{2} + 12 x - 14 y + 36 = 0

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