What is 3x^2+3y^2-18x+12y-36=0 ?

The solution to 3x^2+3y^2-18x+12y-36=0 is Circle with (a,b)=(3,-2),r=5

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3x^2+3y^2-18x+12y-36=0

3 x^{2} + 3 y^{2} - 18 x + 12 y - 36 = 0

(a, b) = (3, - 2), r = 5

Solution steps

3 x^{2} + 3 y^{2} - 18 x + 12 y - 36 = 0

Rewrite

3 x^{2} + 3 y^{2} - 18 x + 12 y - 36 = 0

in the form of the standard circle equation

(x - 3)^{2} + (y - (- 2))^{2} = 5^{2}

Therefore the circle properties are:

(a, b) = (3, - 2), r = 5

(x - 4)^{2} + (y + 1)^{2} = 25

(x - 4)^{2} + (y + 1)^{2} = 16

(x + 1)^{2} + (y + 5)^{2} = 25

x^{2} + y^{2} + y = 9

(x + 2)^{2} + (y - 3)^{2} = 5

What is 3x^2+3y^2-18x+12y-36=0 ?
The solution to 3x^2+3y^2-18x+12y-36=0 is Circle with (a,b)=(3,-2),r=5