Question 583783
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Use the idea that the perpendicular bisector of any chord of a circle passes through the center of the circle.  Since the given points are on the circle, any pair of the three points are the endpoints of a line segment that is a chord of the circle.


Select 2 of the three given points.


Step 1:  Calculate the midpoint of the segment connecting those two points using the midpoint formulas:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_m\ = \frac{x_1 + x_2}{2}] and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y_m\ = \frac{y_1 + y_2}{2}]


Step 2:  Calculate the slope of the line containing the two selected points using the slope formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ \frac{y_1\ -\ y_2}{x_1\ -\ x_2} ]


Step 3:  Determine the slope of the perpendicular to the segment.  Take the negative reciprocal of the slope calculated in step 2.


Step 4:  Derive an equation of the perpendicular bisector of the selected segment by using the point-slope form of an equation, the slope from step 3, and the midpoint from step 1.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ m(x\ -\ x_1) ]


Choose a different pair of points and repeat the process.


Take the two equations of the perpendicular bisectors as a 2X2 system and solve for the point of intersection, *[tex \LARGE (h,\,k)] which is the center of the circle.


Use a modified form of the distance formula to calculate the radius squared:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r^2\ =\ (x_1\ -\ h)^2\ +\ (y_1\ -\ k)^2]


where *[tex \LARGE (x_1,y_1)] is any of the initially given points.


Using the coordinates of the center and the radius, write the standard form equation:


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


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