What is x^2+4x+y^2-3y-18=0 ?

The solution to x^2+4x+y^2-3y-18=0 is Circle with (a,b)=(-2, 3/2),r=(sqrt(97))/2

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x^2+4x+y^2-3y-18=0

x^{2} + 4 x + y^{2} - 3 y - 18 = 0

(a, b) = (- 2, \frac{3}{2}), r = \frac{97}{2}

Solution steps

x^{2} + 4 x + y^{2} - 3 y - 18 = 0

Rewrite

x^{2} + 4 x + y^{2} - 3 y - 18 = 0

in the form of the standard circle equation

(x - (- 2))^{2} + (y - \frac{3}{2})^{2} = (\frac{97}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 2, \frac{3}{2}), r = \frac{97}{2}

x^{2} + 14 x + y^{2} - 2 y + 25 = 0

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x^{2} + y^{2} - 26 x + 2 y + 120 = 0

x^{2} + y^{2} - 8 x - 8 (2 x + 1) + 27 = 0

(x - 4)^{2} + (y - 7)^{2} = (13)^{2}

What is x^2+4x+y^2-3y-18=0 ?
The solution to x^2+4x+y^2-3y-18=0 is Circle with (a,b)=(-2, 3/2),r=(sqrt(97))/2