SOLUTION: Sketch the graph of x^2+y^2+2x-6y-6=0. I believe I have to use the hyperbola formula but I don't know how to do it.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Sketch the graph of x^2+y^2+2x-6y-6=0. I believe I have to use the hyperbola formula but I don't know how to do it.      Log On


   

Question 547019: Sketch the graph of x^2+y^2+2x-6y-6=0.
I believe I have to use the hyperbola formula but I don't know how to do it.
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given to graph:
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x%5E2%2By%5E2%2B2x-6y-6=0
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Rearrange the terms containing x and the terms containing y as follows:
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x%5E2%2B2x%2By%5E2-6y-6=0
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Add 6 to both sides to get rid of the -6 on the left side. Doing this results in:
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x%5E2%2B2x+%2By%5E2-6y=6
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Add +1 and +9 to both sides:
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x%5E2+%2B2x+%2B1+%2By%5E2+-6y+%2B+9+=+6+%2B1%2B9
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You got the +1 by dividing the coefficient of the x term (+2) by 2 to get +1 and then squared the +1 to get +1 which you then added to both sides. You got the +9 by dividing the coefficient of the y term (-6) by 2 to get -3 and then squared that to get +9 which you then added to both sides.
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The left side now contains two tri-nomials, one in x and one in y as grouped below:
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%28x%5E2+%2B2x+%2B1%29+%2B+%28y%5E2+-6y+%2B+9%29+=+6+%2B1%2B9
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Note that on the each of the two groups on the left side are perfect squares as shown:
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%28x%2B1%29%5E2+%2B+%28y+-+3%29%5E2+=+6+%2B+1+%2B+9
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Add the three constants on the right side:
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%28x%2B1%29%5E2+%2B+%28y+-+3%29%5E2+=+16
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Note that the constant on the right side is equal to 4%5E2. Substitute this in for the right side and you have:
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%28x%2B1%29%5E2+%2B+%28y+-+3%29%5E2+=+4%5E2
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This should now be a form that you recognize. It is a circle centered at (-1,3) and having a radius equal to 4.
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Hope this helps you to understand a method of working with equations such as this. We just did some additions of constants to form two perfect squares ... using the method for completing the square as you probably did at one time when you were learning how the quadratic formula was developed.
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