Question

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

Answer 1

Answer:

Center of the circle: $(-1,\frac{1}{6})$

Radius of the circle: $\frac{\sqrt{37} }{6}$

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:

$3x^2+3y^2+6x-y=0\\x^2+y^2+2x-\frac{1}{3} y=0$

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":

$x^2+y^2+2x-\frac{1}{3} y=0\\( x^2+2x ) + (y^2-\frac{1}{3} y)=0$

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":

$( x^2+2x ) = 0\\x^2+2x+1=1\\(x+1)^2=1$

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":

$(y^2-\frac{1}{3} y)=0\\y^2-\frac{1}{3} y+\frac{1}{36} =\frac{1}{36}\\(y-\frac{1}{6} )^2=\frac{1}{36}$

Therefore, we need to add "$\frac{1}{36}$" to both sides to complete the square for the y-variable.

This means we need to add a total of   $1 + \frac{1}{36} = \frac{37}{36}$ to both sides of the initial equation in order to complete the square for both variables:

$x^2+y^2+2x-\frac{1}{3} y=0\\x^2+y^2+2x-\frac{1}{3} y+\frac{37}{36} =\frac{37}{36} \\(x+1)^2+(y-\frac{1}{6} )^2=\frac{37}{36}$

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 $(x_0,y_0)$  and radius R:

$(x-x_0)^2+(y-y_0)^2=R^2$

So our equation that can be written as:

$(x+1)^2+(y-\frac{1}{6} )^2=(\sqrt{\frac{37}{36}} )^2$

corresponds to a circle centered at   $(-1,\frac{1}{6})$  , and with radius $\sqrt{\frac{37}{36}}=\frac{\sqrt{37} }{6}$