Question 422050
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Complete the square:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ y^2\ +\ 4x\ -\ 6y\ -\ 3\ =\ 0]


Constant term to the RHS:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ y^2\ +\ 4x\ -\ 6y\ =\ 3]


Group like variables:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 4x\ +\ y^2\ -\ 6y\ =\ 3]


Divide the first degree x term coefficient by 2, square the result, and add that result to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 4x\ +\ 4\ +\ y^2\ -\ 6y\ =\ 3\ +\ 4]


Repeat for the first degree y term:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 4x\ +\ 4\ +\ y^2\ -\ 6y\ +\ 9\ =\ 3\ +\ 4\ +\ 9]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 2)^2\ +\ (y\ -\ 3)^2\ =\ 16]


The equation of a circle with center at *[tex \Large (h,k)] and radius *[tex \Large r] is:


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


So re-write your equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ -\ (-2)\right)^2\ +\ \left(y\ -\ 3\right)^2\ =\ 4^2]



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