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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

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

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

x^{2} + 5 x + y^{2} - 7 y + \frac{247}{16} = 0

(x - 5)^{(2)} + (y + 7)^{(2)} = 50

x^{2} - 16 x + 64 + y^{2} - 16 x = 0

x^{2} + y^{2} + 6 x - 4 y - 4 = 0

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