What is (x^2+y^2-2x+y-3)/2 >0 ?

The solution to (x^2+y^2-2x+y-3)/2 >0 is Circle with (a,b)=(1,-1/2),r=(sqrt(17))/2

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(x^2+y^2-2x+y-3)/2 >0

\frac{x ^{2} + y ^{2} - 2 x + y - 3}{2} > 0

(a, b) = (1, - \frac{1}{2}), r = \frac{17}{2}

Solution steps

\frac{x ^{2} + y ^{2} - 2 x + y - 3}{2} > 0

Rewrite

\frac{x ^{2} + y ^{2} - 2 x + y - 3}{2} > 0

in the form of the standard circle equation

(x - 1)^{2} + (y - (- \frac{1}{2}))^{2} = (\frac{17}{2})^{2}

Therefore the circle properties are:

(a, b) = (1, - \frac{1}{2}), r = \frac{17}{2}

x^{2} + y^{2} = 53

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

(x + 3)^{2} + (y - 2)^{2} = \frac{25}{4}

x^{2} + y^{2} = 13

x^{2} + y^{2} + 6 x + 5 y = \frac{45}{4}

What is (x^2+y^2-2x+y-3)/2 >0 ?
The solution to (x^2+y^2-2x+y-3)/2 >0 is Circle with (a,b)=(1,-1/2),r=(sqrt(17))/2