What is x^2+y^2-3/2 x+2/3 y-239/36 =0 ?

The solution to x^2+y^2-3/2 x+2/3 y-239/36 =0 is Circle with (a,b)=(3/4 ,-1/3),r=(3sqrt(13))/4

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x^2+y^2-3/2 x+2/3 y-239/36 =0

x^{2} + y^{2} - \frac{3}{2} x + \frac{2}{3} y - \frac{239}{36} = 0

(a, b) = (\frac{3}{4}, - \frac{1}{3}), r = \frac{3 13}{4}

Solution steps

x^{2} + y^{2} - \frac{3}{2} x + \frac{2}{3} y - \frac{239}{36} = 0

Rewrite

x^{2} + y^{2} - \frac{3}{2} x + \frac{2}{3} y - \frac{239}{36} = 0

in the form of the standard circle equation

(x - \frac{3}{4})^{2} + (y - (- \frac{1}{3}))^{2} = (\frac{3 13}{4})^{2}

Therefore the circle properties are:

(a, b) = (\frac{3}{4}, - \frac{1}{3}), r = \frac{3 13}{4}

[x - (- 3)]^{2} + (y - 1)^{2} = 1 0^{2}

x^{2} + y^{2} + 18 x = - 80

x^{2} + y^{2} > \frac{1}{2}

x^{2} - 8 x + y^{2} - 4 y + 10 = 0

x^{2} + x + y^{2} = 16

What is x^2+y^2-3/2 x+2/3 y-239/36 =0 ?
The solution to x^2+y^2-3/2 x+2/3 y-239/36 =0 is Circle with (a,b)=(3/4 ,-1/3),r=(3sqrt(13))/4