What is x^2-3x+y^2+2y-14=0 ?

The solution to x^2-3x+y^2+2y-14=0 is Circle with (a,b)=(3/2 ,-1),r=(sqrt(69))/2

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x^2-3x+y^2+2y-14=0

x^{2} - 3 x + y^{2} + 2 y - 14 = 0

(a, b) = (\frac{3}{2}, - 1), r = \frac{69}{2}

Solution steps

x^{2} - 3 x + y^{2} + 2 y - 14 = 0

Rewrite

x^{2} - 3 x + y^{2} + 2 y - 14 = 0

in the form of the standard circle equation

(x - \frac{3}{2})^{2} + (y - (- 1))^{2} = (\frac{69}{2})^{2}

Therefore the circle properties are:

(a, b) = (\frac{3}{2}, - 1), r = \frac{69}{2}

x^{2} + y^{2} + 14 x - 8 y + 56 = 0

(x - 5)^{2} + (y + 2)^{2} = 49

x^{2} + y^{2} = 625

(x + 3) (x + 4) + (y - 4) (y - 3) = 0

(x + 5)^{2} + (y + 8)^{2} = 20

What is x^2-3x+y^2+2y-14=0 ?
The solution to x^2-3x+y^2+2y-14=0 is Circle with (a,b)=(3/2 ,-1),r=(sqrt(69))/2