Question 608352
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ y^2\ +\ 18x\ +\ 8y\ =\ -93] 


Rearrange:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 18x\ +\ y^2\ +\ 8y\ =\ -93]


Divide the coefficient on the 1st degree *[tex \LARGE x] term by 2.  Square the result.  Add the square to both sides of the equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 18x\ +\ 81\ +\ y^2\ +\ 8y\ =\ -93\ +\ 81]


Divide the coefficient on the 1st degree *[tex \LARGE y] term by 2.  Square the result.  Add the square to both sides of the equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 18x\ +\ 81\ +\ y^2\ +\ 8y\ +\ 16\ =\ -93\ +\ 81\ +\ 16]


Factor the two perfect square trinomials in the LHS and collect terms in the RHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 9)^2\ +\ (y\ +\ 4)^2\ =\ 4]


Rewrite for standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ (-9)\right)^2\ +\ \left(y\ -\ (-4)\right)^2\ =\ 2^2]


A circle with center at *[tex \LARGE (h, k)] and radius *[tex \LARGE r] has an equation:


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


So you should be able to determine the center coordinates and radius by inspection.


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