show that the equation 3x2 + 3y2 + 6x-y= 0 represents a circle and find the centre and radius of the circle

Center of the circle:

Radius of the circle:

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:

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":

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":

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":

Therefore, we need to add "" to both sides to complete the square for the y-variable.

This means we need to add a total of   to both sides of the initial equation in order to complete the square for both variables:

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  and radius R:

So our equation that can be written as:

corresponds to a circle centered at    , and with radius

based on context clues you gave i'm assuming the answer would be c.

a is the correct