Question 544659
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All points a given distance from a given point is a circle where the given point is the center.  The equation of a circle with center at the point *[tex \Large \left(h,k\right)] and radius *[tex \Large r] is *[tex \Large (x\ -\ h)^2\ (y\ -\ k)^2\ =\ r^2]


So any point where the coordinates of the ordered pair identifying that point satisfy


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


is, by definition of a circle, the radius, i.e. *[tex \Large \sqrt{2}], distant from the given point (-1, 5).


There are two ways to proceed.  You can pick any value you like for either *[tex \Large x] in the interval *[tex \Large -1\ -\ \sqrt{2}\ \leq\ x\ \leq\ -1\ +\ \sqrt{2}] or *[tex \Large y] in the interval *[tex \Large 5\ -\ \sqrt{2}\ \leq\ y\ \leq\ 5\ +\ \sqrt{2}], substitute that value, and then solve for the value of the other variable, giving you an ordered pair that is on the circle and the proper distance from the given point.


Or you can do it the easy way:  Replace the coordinate -1 in the given point with either *[tex \Large -1\ +\ \sqrt{2}] or *[tex \Large -1\ -\ \sqrt{2}] OR replace the coordinate 5 in the given point with either *[tex \Large 5\ +\ \sqrt{2}] or *[tex \Large 5\ -\ \sqrt{2}]. 


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