Question 800665
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Use the fact that the perpendicular bisector of a chord of a circle passes through the center of the circle.


Label your points A, B, and C.  Doesn't matter which is which.  Draw segments AB and BC


Use the slope formula to find the slope of the line containing segment AB.


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


where *[tex \Large \left(x_1,y_1\right)] and *[tex \Large \left(x_2,y_2\right)] are the coordinates of the given points.


Then do the same thing for the line containing segment BC.


Use the midpoint formulas to calculate the midpoints of segments AB and BC.


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


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


Use the negative reciprocal of the slope of the line containing segment AB and the coordinates of the midpoint of segment AB in the point-slope form of an equation of a line to derive an equation of the perpendicular bisector of segment AB.


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


where *[tex \Large \left(x_1,y_1\right)] are the coordinates of the given point and *[tex \Large m] is the given/calculated slope.


Repeat to find an equation of the perpendicular bisector of segment BC.


Use any convenient method to solve the 2X2 system comprised of the perpendicular bisector equations you just derived.  The point of intersection will be the center of the circle.


Use the distance formula with the center and any one of the three given points to determine the radius of the circle.  You can use the modified form of the distance formula shown below because you actually need the radius squared to complete the circle equation.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d^2\ =\ r^2\ =\ (x_1\ -\ x_2)^2\ +\ (y_1\ -\ y_2)^2]


Now that you have the coordinates of the center of the circle, *[tex \Large (h,\,k)] and the radius squared *[tex \Large r^2], you can write the equation of the circle directly, just fill in the numbers.


The equation of a circle with center at *[tex \Large \left(h,\,k\right)] and radius *[tex \Large r] is *[tex \Large \left(x\ -\ h\right)^2\ +\ \left(y\ -\ k\right)^2\ =\ r^2].


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \