Question 376355
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The equation of a circle centered at *[tex \Large (h, k)] with radius *[tex \Large r] is:


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


Since your circles are all centered at the origin, you have:


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


Recalling Pythagoras, if *[tex \Large (x_1,y_1)] is a point on the circle, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r^2\ =\ x_1^2\ +\ y_1^2]


Once you have the value of *[tex \Large r^2] you can add the opposite of that value to both sides so that you have:


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


Which is equivalent to Standard Form: *[tex \Large Ax^2\ +\  By^2\ +\ Cx\ +\ Dy\ +\ E\ =\ 0] where A = 1,  B = 1, C and D = 0, and E = *[tex \Large -r^2]



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