What is 3x^2+3y^2+18x-9y-12=0 ?

The solution to 3x^2+3y^2+18x-9y-12=0 is Circle with (a,b)=(-3, 3/2),r=(sqrt(61))/2

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3x^2+3y^2+18x-9y-12=0

3 x^{2} + 3 y^{2} + 18 x - 9 y - 12 = 0

(a, b) = (- 3, \frac{3}{2}), r = \frac{61}{2}

Solution steps

3 x^{2} + 3 y^{2} + 18 x - 9 y - 12 = 0

Rewrite

3 x^{2} + 3 y^{2} + 18 x - 9 y - 12 = 0

in the form of the standard circle equation

(x - (- 3))^{2} + (y - \frac{3}{2})^{2} = (\frac{61}{2})^{2}

Therefore the circle properties are:

(a, b) = (- 3, \frac{3}{2}), r = \frac{61}{2}

x^{2} + y^{2} - 14 x - 12 y + 59 = 0

2 x^{2} + 2 y^{2} - 4 x + y + 1 = 0

x^{2} + 14 x + y^{2} + 10 y + 62 = 0

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

x^{2} + y^{2} - 2 x - 6 y + 7 = 0

What is 3x^2+3y^2+18x-9y-12=0 ?
The solution to 3x^2+3y^2+18x-9y-12=0 is Circle with (a,b)=(-3, 3/2),r=(sqrt(61))/2