Question 244126
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You need to complete the square on both *[tex \Large x] and *[tex \Large y], then factor the perfect squares.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ y^2\ +\ 8x\ -\ 2y\ +\ 15\ =\ 0]


First, move the constant to the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ y^2\ +\ 8x\ -\ 2y\ =\ -15]


Rearrange:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 8x\ +\ y^2\ -\ 2y\ =\ -15]


Divide the coefficient on the *[tex \Large x]-term by 2, then square the result.  Add this result to both sides of the equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 8x\ +\ 16\ +\ y^2\ -\ 2y\ =\ -15\ +\ 16]


Do the same thing for the *[tex \Large y]-term:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 8x\ +\ 16\ +\ y^2\ -\ 2y\ +\ 1\ =\ -15\ +\ 16\ +\ 1\ =\ 2]


Factor *[tex \LARGE x^2\ +\ 8x\ + 16] and *[tex \LARGE y^2\ -\ 2y\ +\ 1] and then replace these two trinomials in your equation with the squared binomial representation.


You can then determine the center by inspection since a circle centered at *[tex \Large \left(h,k\right)] with radius *[tex \Large r] is defined by:


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


Pay careful attention to the signs on your center coordinates.


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