document.write( "Question 1079001: Find the equation of the circle that passes through (1,2) and (3,4); centre on 3x + y - 13 = 0. \n" ); document.write( "

Algebra.Com's Answer #693378 by ikleyn(52812) $\"\"$ $\"About$

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document.write( "The locus of the points equidistant from two given points is the perpendicular bisector to the segment connecting these points.\r\n" );
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document.write( "So, you need to construct the perpendicular bisector to the segment connecting (1,2) and (3,4), and then \r\n" );
document.write( "find its intersection with the given straight line.\r\n" );
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document.write( "The midpoint between the two given points is (2,3).\r\n" );
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document.write( "The segment connecting (1,2) and (3,4) has the slope  = 1.\r\n" );
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document.write( "Hence, the perpendicular line (perpendicular bisector) has the slope -1.\r\n" );
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document.write( "The line with the slope -1 passing through the point (2,3) has the equation \r\n" );
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document.write( "y - 3 = (-1)*(x-2),   or\r\n" );
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document.write( "y = -x + 5.\r\n" );
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document.write( "The intersection of the straight lines\r\n" );
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document.write( "3x + y = 13     (1)   (the given line)   and\r\n" );
document.write( "y = -x + 5      (2)   (the perpendicular bisector)\r\n" );
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document.write( "is (solve the system by substitution) the point (x,y) = (4,1)\r\n" );
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document.write( "So, the center of the circle is the point (4,1).\r\n" );
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document.write( "The radius of the circle is the distance from the point (4,1) to the point (3,4):\r\n" );
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document.write( "r =  =  = .\r\n" );
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document.write( "Therefore, the equation of this circle is \r\n" );
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document.write( " = 10.\r\n" );
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