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MrMuchimi
3 years ago
10

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

Mathematics
1 answer:
Jlenok [28]3 years ago
5 0

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}

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