What is x^2+y^2-x+3y-5=0 ?

The solution to x^2+y^2-x+3y-5=0 is Circle with (a,b)=(1/2 ,-3/2),r=sqrt(15/2)

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x^2+y^2-x+3y-5=0

x^{2} + y^{2} - x + 3 y - 5 = 0

(a, b) = (\frac{1}{2}, - \frac{3}{2}), r = \frac{15}{2}

Solution steps

x^{2} + y^{2} - x + 3 y - 5 = 0

Rewrite

x^{2} + y^{2} - x + 3 y - 5 = 0

in the form of the standard circle equation

(x - \frac{1}{2})^{2} + (y - (- \frac{3}{2}))^{2} = (\frac{15}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, - \frac{3}{2}), r = \frac{15}{2}

x^{2} + y^{2} \geq 16

x^{2} + y^{2} - 4 x + 14 y + 4 = 0

x^{2} + y^{2} - 1 > 0

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

3 x^{2} + 3 y^{2} - 12 x + 6 y - 12 = 0

What is x^2+y^2-x+3y-5=0 ?
The solution to x^2+y^2-x+3y-5=0 is Circle with (a,b)=(1/2 ,-3/2),r=sqrt(15/2)