document.write( "Question 388559: standard points of a circle using the point s (-6,3) (-4,-1)(-2,5)
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Algebra.Com's Answer #274902 by robertb(5830) $\"\"$ $\"About$

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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 $\"y+=+%28x%2B5%29%2F2+%2B+1\"$ .
\n" ); document.write( "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 $\"y+=+-%28x%2B3%29%2F3+%2B+2\"$ .
\n" ); document.write( "The intersection of the two perpendicular bisectors is the center of the circle.
\n" ); document.write( " $\"%28x%2B5%29%2F2+%2B+1%7D+=+-%28x%2B3%29%2F3+%2B+2\"$ . Multiplying both sides by 6, we get $\"3x%2B15+%2B+6+=+-2x-6%2B12\"$ , 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. \n" ); document.write( "

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