Question 385641
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Complete the square on each of the variables:


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


Group variable types:


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


Start with the x term.  Divide the coefficient by 2, square the result, add the squared result to both sides of the equation.  -2 divided by 2 is -1, -1 squared is 1, add 1 to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x^2\ -\ 2x\ +\ 1\right)\ +\ \left(y^2\ +\ 4y\right)\ =\ 4\ +\ 1]


Repeat the process for y:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x^2\ -\ 2x\ +\ 1\right)\ +\ \left(y^2\ +\ 4y\ +\ 4\right)\ =\ 4\ +\ 1\ +\ 4]


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


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


The constant term in the RHS is now the radius squared.  The radius is the positive square root of the constant.


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