Question 516646
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ h)^2\ +\ (y\ -\ k)^2\ =\ r^2]


is the equation of a circle centered at *[tex \Large (h,k)] with radius *[tex \Large r]


So:


Divide through by 5


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ y^2\ +\ 6x\ -\ 4y\ -\ 12\ =\ 0]


Move the constant into the RHS and put the x-terms together and the y-terms together:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 6x\ +\ y^2\ -\ 4y\ =\ 12]


Divide the coefficient on the first degree x-term by 2, square the result, add that result to both sides of the equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 6x\ +\ 9\ +\ y^2\ -\ 4y\ =\ 12\ +\ 9]


Do the same thing with the coefficient on the first degree y-term:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 6x\ +\ 9\ +\ y^2\ -\ 4y\ +\ 4\ =\ 12\ +\ 9\ +\ 4\ =\ 25]


Factor the two perfect square trinomials in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 3)^2\ +\ (y\ -\ 2)^2\ =\ 25]


Now everything you need to know is right there in front of you.  Hint:  Rewrite your equation thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ -\ (-3))^2\ +\ (y\ -\ 2)^2\ =\ 5^2]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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