What is 3x^2+3y^2+8x+6y+7=0 ?

The solution to 3x^2+3y^2+8x+6y+7=0 is Circle with (a,b)=(-4/3 ,-1),r= 2/3

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3x^2+3y^2+8x+6y+7=0

3 x^{2} + 3 y^{2} + 8 x + 6 y + 7 = 0

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

Solution steps

3 x^{2} + 3 y^{2} + 8 x + 6 y + 7 = 0

Rewrite

3 x^{2} + 3 y^{2} + 8 x + 6 y + 7 = 0

in the form of the standard circle equation

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

Therefore the circle properties are:

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

36 x^{2} + 36 y^{2} - 48 x - 36 y + 16 = 0

3 x^{2} + 3 y^{2} - 15 - 0 = 0

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

(x - 6)^{2} + (y - (- 2))^{2} = 25

x^{2} + y^{2} - 8 x - 1 y - 74 = 0

What is 3x^2+3y^2+8x+6y+7=0 ?
The solution to 3x^2+3y^2+8x+6y+7=0 is Circle with (a,b)=(-4/3 ,-1),r= 2/3