Question 403004
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You are right -- that is not right.  Why did you add 16 and 4 to the RHS when you only added 4 and 1 to the LHS?  There is no factor of 4 in front of either set of parentheses in the LHS.  Furthermore, if 40 <i>was</i> the correct value in the RHS, then the RHS would be properly expressed as *[tex \Large \left(\sqrt{40}\right)^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x^2\ +\ 4x\ +\ 4\right)\ +\ \left(y^2\ -\ 2y\ +\ 1\right)\ =\ 20\ +\ 4\ +\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ +\ 2\right)^2\ +\ \left(y\ -\ 1\right)^2\ =\ (5)^2]


Center at *[tex \Large (-2,1)], radius *[tex \Large 5]


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