Question 388559: standard points of a circle using the point s (-6,3) (-4,-1)(-2,5)
center?
radius?
Answer by robertb(5830)   (Show Source):
You can put this solution on YOUR website! The midpoint of (-6, 3) and (-4,-1) is (-5, 1). The slope of the line passing through (-6, 3) and (-4,-1) is (3--1)/(-6--4) = 4/-2 = -2. The equation of the line passing through the midpoint and perpendicular to the line through (-6, 3) and (-4,-1) is y-1 = (x--5)/2, or  .
The midpoint of (-2,5) and (-4,-1) is (-3, 2). The slope of the line passing through (-2, 5) and (-4,-1) is (5--1)/(-2--4) = 6/2 = 3. The equation of the line passing through the midpoint and perpendicular to the line through (-2, 5) and (-4,-1) is y-2 = -(x--3)/3, or  .
The intersection of the two perpendicular bisectors is the center of the circle.
 . Multiplying both sides by 6, we get  , or 3x+21 = -2x + 6, or 5x = -15, or x = -3. To solve for y substitute into any one of the equations for the perpendicular bisector. The radius is obtained by getting the distance of the center from any one of the given points.
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