Question 1119612
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Principles:


The perpendicular bisectors of any chords of a circle intersect in the center of the circle.


The slope of a perpendicular bisector is the negative reciprocal of the slope of the original line.


The distance from the center to any of the three given points is the radius.


Procedure:


Select any pair of the given points, *[tex \Large (x_1,y_1)] and *[tex \Large (x_2,y_2)], and use the slope formula to calculate the slope of the line containing those two points.


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


Then calculate the negative reciprocal of this slope, i.e. *[tex \Large -\frac{1}{m}]


Use the midpoint formulas to calculate the coordinates of the midpoint, *[tex \Large (x_m,\,y_m)] of the chord segment:


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


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


Derive the slope-intercept form of an equation of the line that has the negative reciprocal slope and passes through the calculated midpoint, thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y\ =\ -\frac{1}{m}(x\ -\ x_m)\ +\ y_m]


Repeat the above process for a different pair of the given points.


Take the RHS of each of the derived equations and set them equal to each other.  Solve for *[tex \Large x] to obtain the *[tex \Large x]-coordinate of the center of the desired circle.  Substitute this value and calculate the *[tex \Large y]-coordinate of the center of the circle.  You now have determined the circle center, *[tex \Large (h,\,k)]


Use the distance formula with any one of the given points and the center, *[tex \Large (h,\,k)] to calculate the measure of the radius, *[tex \Large r].  Since you will ultimately require only the square of the radius, you need not do the square root step in the distance formula.


Finally, construct the Standard Form:


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


And then expand the binomials and collect like terms to create the General Form as required.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^2\ +\ y^2\ +\ Dx\ +\ Ey\ +\ F\ =\ 0]


Where *[tex \Large D\ =\ -2h,\ \ ]*[tex \Large E\ =\ -2k,\ \ ]and*[tex \Large \ F\ =\ h^2\ +\ 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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*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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