Question 198055
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First, put it in general form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 6x + y^2 + 2y = -1]


You have both an x-squared term and a y-squared term, so you eliminate parabola as the conic type.


The signs on these terms are the same so you eliminate hyperbola.


The coefficients on the high-order terms are equal, so you eliminate ellipse.


This is a circle.


Complete the square on x:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 6x + y^2 + 2y = -1]


-6 divided by 2 is -3, squared is 9 -- add 9 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 6x + 9 + y^2 + 2y = -1 + 9]


Complete the square on y:


2 divided by 2 is 1, squared is 1 -- add 1 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2 - 6x + 9 + y^2 + 2y + 1 = -1 + 9 + 1 = 9]


Factor the two perfect squares:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x - 3)^2 + (y + 1)^2 = 9]


Center at (3,-1), radius 3.  You can draw your own graph.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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