What is x^2+y^2+2x+10/3 y-16=0 ?

The solution to x^2+y^2+2x+10/3 y-16=0 is Circle with (a,b)=(-1,-5/3),r=(sqrt(178))/3

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x^2+y^2+2x+10/3 y-16=0

x^{2} + y^{2} + 2 x + \frac{10}{3} y - 16 = 0

(a, b) = (- 1, - \frac{5}{3}), r = \frac{178}{3}

Solution steps

x^{2} + y^{2} + 2 x + \frac{10}{3} y - 16 = 0

Rewrite

x^{2} + y^{2} + 2 x + \frac{10}{3} y - 16 = 0

in the form of the standard circle equation

(x - (- 1))^{2} + (y - (- \frac{5}{3}))^{2} = (\frac{178}{3})^{2}

Therefore the circle properties are:

(a, b) = (- 1, - \frac{5}{3}), r = \frac{178}{3}

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

(x - 8)^{2} + (y + 3)^{2} = 91

x^{2} + y^{2} - 12 x - 10 y - 15 = 0

\frac{3}{2} x^{2} + \frac{3}{2} y^{2} = \frac{75}{2}

16 x^{2} + 16 y^{2} + 8 x - 32 = - 1

What is x^2+y^2+2x+10/3 y-16=0 ?
The solution to x^2+y^2+2x+10/3 y-16=0 is Circle with (a,b)=(-1,-5/3),r=(sqrt(178))/3