Answer:
Center of the circle: [tex](-1,\frac{1}{6})[/tex]
Radius of the circle: [tex]\frac{\sqrt{37} }{6}[/tex]
Step-by-step explanation:
Let's start by dividing both sides of the equation by the factor "3" so we simplify our next step of completing squares for x and for y:
[tex]3x^2+3y^2+6x-y=0\\x^2+y^2+2x-\frac{1}{3} y=0[/tex]
Now we work on completing the squares for the expression on x and for the expression on y separately, so we group together the terms in "x" and then the terms in "y":
[tex]x^2+y^2+2x-\frac{1}{3} y=0\\( x^2+2x ) + (y^2-\frac{1}{3} y)=0[/tex]
Let's find what number we need to add to both sides of the equation to complete the square of the group on the variable "x":
[tex]( x^2+2x ) = 0\\x^2+2x+1=1\\(x+1)^2=1[/tex]
So, we need to add "1" to both sides in order to complete the square in "x".
Now let's work on a similar fashion to find what number we need to add on both sides to complete the square for the group on y":
[tex](y^2-\frac{1}{3} y)=0\\y^2-\frac{1}{3} y+\frac{1}{36} =\frac{1}{36}\\(y-\frac{1}{6} )^2=\frac{1}{36}[/tex]
Therefore, we need to add "[tex]\frac{1}{36}[/tex]" to both sides to complete the square for the y-variable.
This means we need to add a total of [tex]1 + \frac{1}{36} = \frac{37}{36}[/tex] to both sides of the initial equation in order to complete the square for both variables:
[tex]x^2+y^2+2x-\frac{1}{3} y=0\\x^2+y^2+2x-\frac{1}{3} y+\frac{37}{36} =\frac{37}{36} \\(x+1)^2+(y-\frac{1}{6} )^2=\frac{37}{36}[/tex]
Now recall that the right hand side of this expression for the equation of a circle contains the square of the circle's radius, based on the general form for the equation of a circle of center [tex](x_0,y_0)[/tex] and radius R:
[tex](x-x_0)^2+(y-y_0)^2=R^2[/tex]
So our equation that can be written as:
[tex](x+1)^2+(y-\frac{1}{6} )^2=(\sqrt{\frac{37}{36}} )^2[/tex]
corresponds to a circle centered at [tex](-1,\frac{1}{6})[/tex] , and with radius [tex]\sqrt{\frac{37}{36}}=\frac{\sqrt{37} }{6}[/tex]
