document.write( "Question 990592: Find the equation of the circle passing through the points A(3,2),B(5,3) and C(2,5) \n" ); document.write( "
Algebra.Com's Answer #610663 by Edwin McCravy(20054)\"\" \"About 
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document.write( "The easiest way is to use the fact that the perpendicular bisector of a chord\r\n" );
document.write( "is an extended diameter. Then two perpendicular bisectors of chords must\r\n" );
document.write( "intersect at the center of the circle.  Then since to have the center you will\r\n" );
document.write( "be able to find the radius with the distance formula.\r\n" );
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document.write( "1. Find the slope of chord AB\r\n" );
document.write( "2. Find the midpoint of chord AB\r\n" );
document.write( "3. Find the equation of the perpendicular bisector of chord AB.\r\n" );
document.write( "   (a) Use slope which is the negative reciprocal of the slope of AB,\r\n" );
document.write( "   (b) Use midpoint of AB, \r\n" );
document.write( "   (c) Use the point-slope formula. \r\n" );
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document.write( "4. Find the slope of either chord BC [or AC].\r\n" );
document.write( "5. Find the midpoint of chord BC [or AC]\r\n" );
document.write( "6. Find the equation of the perpendicular bisector of chord BC [or AC]\r\n" );
document.write( "7. Solve the system of two equations obtained in steps 3 and 6.\r\n" );
document.write( "8. The solution to 7 will be the center, (h,k)\r\n" );
document.write( "9. Use the distance formula to find r, the radius which is the distance \r\n" );
document.write( "   from the center to any of the three given points/\r\n" );
document.write( "10. Then substitute h, k, and r in this general form for the equation of a circle.:\r\n" );
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document.write( "      (x-h)2 + (y-k)2 = r2   \r\n" );
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document.write( "The other way, which is harder, is to substitute the three points into that\r\n" );
document.write( "equation and get a system of three equations in h, k and r, and solve for those\r\n" );
document.write( "variables.\r\n" );
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document.write( "      (x-h)2 + (y-k)2 = r2\r\n" );
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document.write( "Edwin

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