What is x^2+y^2+3x-8y+18=0 ?

The solution to x^2+y^2+3x-8y+18=0 is Circle with (a,b)=(-3/2 ,4),r= 1/2

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x^2+y^2+3x-8y+18=0

x^{2} + y^{2} + 3 x - 8 y + 18 = 0

(a, b) = (- \frac{3}{2}, 4), r = \frac{1}{2}

Solution steps

x^{2} + y^{2} + 3 x - 8 y + 18 = 0

Rewrite

x^{2} + y^{2} + 3 x - 8 y + 18 = 0

in the form of the standard circle equation

(x - (- \frac{3}{2}))^{2} + (y - 4)^{2} = (\frac{1}{2})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{3}{2}, 4), r = \frac{1}{2}

x^{2} + y^{2} - 2 x + 28 y + 181 = 0

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

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

x^{2} + y^{2} + 4 x + 6 y - 3 = 0

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

What is x^2+y^2+3x-8y+18=0 ?
The solution to x^2+y^2+3x-8y+18=0 is Circle with (a,b)=(-3/2 ,4),r= 1/2