What is 3x^2+3y^2+5x-6y+1=0 ?

The solution to 3x^2+3y^2+5x-6y+1=0 is Circle with (a,b)=(-5/6 ,1),r= 7/6

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3x^2+3y^2+5x-6y+1=0

3 x^{2} + 3 y^{2} + 5 x - 6 y + 1 = 0

(a, b) = (- \frac{5}{6}, 1), r = \frac{7}{6}

Solution steps

3 x^{2} + 3 y^{2} + 5 x - 6 y + 1 = 0

Rewrite

3 x^{2} + 3 y^{2} + 5 x - 6 y + 1 = 0

in the form of the standard circle equation

(x - (- \frac{5}{6}))^{2} + (y - 1)^{2} = (\frac{7}{6})^{2}

Therefore the circle properties are:

(a, b) = (- \frac{5}{6}, 1), r = \frac{7}{6}

x^{2} - 8 x + y^{2} - 10 y + 36 = 0

x^{2} + y^{2} - 18 x - 12 y + 68 = 0

x^{2} + y^{2} - 6 x - 8 y - 25 = 0

4 x^{2} + 4 y^{2} + 40 x + 32 y = 5

9 x^{2} + 9 y^{2} - 6 x - 12 y - 11 = 0

What is 3x^2+3y^2+5x-6y+1=0 ?
The solution to 3x^2+3y^2+5x-6y+1=0 is Circle with (a,b)=(-5/6 ,1),r= 7/6