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

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

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

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

(a, b) = (3, 2), r = 3

Solution steps

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

Rewrite

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

in the form of the standard circle equation

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

Therefore the circle properties are:

(a, b) = (3, 2), r = 3

(x + 4)^{2} + (y - 6)^{2} = \frac{6}{2}

xx + y^{2} = 5

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

(x - \frac{3}{5})^{2} + (y + \frac{7}{8})^{2} = \frac{9}{4}

x^{2} + y^{2} - 4 x + 5 y - 5 = 0

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