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Question 597818: Classify the conic section and write its equation in standard form.
x^2 + y^2 - 12x + 4y + 31 = 0
Answer by jillfired(3)   (Show Source):
You can put this solution on YOUR website! Classify the conic section and write its equation in standard form.
x^2 + y^2 - 12x + 4y + 31 = 0
This is a circle. To begin you need to move all terms
not containing a variable to the right-hand side of the equation:
x^2-12x+y^2+4y=-31
To create a trinomial square on the left-hand side of the equation, add a value to both sides of the equation that is equal to the square of half the coefficient of x. In this problem, add (-6)^2 (36) to both sides of the equation
x^2-12x+36+y^2+4y=-31+0+36
Factor the perfect trinomial square into (x-6)^2
(x-6)^2+y^2+4y=-31+0+36
To create a trinomial square on the left hand side of the equation, add a value to both sides of the equation that is equal to the square of half the coefficient of y. So in this problem add (2)^2 (4)
(x-6)^2+y^2+4y+4=-31+0+36+0+4
Now just factor the perfect trinomial square into (y+2)^2
(x-6)^2 + (y+2)^2 = -31+0+36+0+4
Add 0 to -31 to get -31
Add 36 to -31 to get 5
(x-6)^2 +(y+2)^2=5+0+0+4
Add 5 and 4 on the right side these up and you come up with
(x-6)^2 + (y+2)^2 = 9 - this is your answer in standard form.
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