What is x^2+y^2-x+3y-15/4 =0 ?

The solution to x^2+y^2-x+3y-15/4 =0 is Circle with (a,b)=(1/2 ,-3/2),r= 5/2

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x^2+y^2-x+3y-15/4 =0

x^{2} + y^{2} - x + 3 y - \frac{15}{4} = 0

(a, b) = (\frac{1}{2}, - \frac{3}{2}), r = \frac{5}{2}

Solution steps

x^{2} + y^{2} - x + 3 y - \frac{15}{4} = 0

Rewrite

x^{2} + y^{2} - x + 3 y - \frac{15}{4} = 0

in the form of the standard circle equation

(x - \frac{1}{2})^{2} + (y - (- \frac{3}{2}))^{2} = (\frac{5}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{1}{2}, - \frac{3}{2}), r = \frac{5}{2}

x^{2} + y^{2} + 6 x - 10 y - 2 = 0

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

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

(x - 8)^{2} + (y + 3)^{2} = 1

x^{2} + y^{2} - 4 x - 12 y - 6 = 0

What is x^2+y^2-x+3y-15/4 =0 ?
The solution to x^2+y^2-x+3y-15/4 =0 is Circle with (a,b)=(1/2 ,-3/2),r= 5/2