document.write( "Question 980426: Can someone please solve these 3 coordinate geometry in the x, y plane questions for me? It would really help. E out. Preferably if you show your working out too so I can see how it's done. \r
\n" ); document.write( "\n" ); document.write( "3) The lines EF and GH are chords of a circle. The line y=3x-24 is the perpendicular bisector of EF. Given G and H are (-2,4) and (4,10) respectively, find the coordinates of the center of the circle. \r
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Algebra.Com's Answer #601569 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "So you'll have something to solve yourself, instead of doing it for you, \r\n" );
document.write( "I'll do one EXACTLY LIKE IT with the numbers changed.  You solve yours\r\n" );
document.write( "exactly like it:\r\n" );
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\n" ); document.write( "3) The lines EF and GH are chords of a circle. The line y=-4x+3 is the
\n" ); document.write( "perpendicular bisector of EF. Given G and H are (1,4) and (2,5) respectively,
\n" ); document.write( "find the coordinates of the center of the circle.
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document.write( "Let (h,k) be the center of the circle.  Then the equation of the circle is \r\n" );
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document.write( "(x-h)² + (y-k)² = r²\r\n" );
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document.write( "Endpoints of a chord of a circle lie on the circle.\r\n" );
document.write( "Therefore (1,4) and (2,5) are points on the circle, and we \r\n" );
document.write( "can substitute them into the equation of the circle and \r\n" );
document.write( "have a true equation:\r\n" );
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document.write( "(1-h)² + (4-k)² = r²\r\n" );
document.write( "(2-h)² + (5-k)² = r²\r\n" );
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document.write( "Since both equal r², we can set them equal:\r\n" );
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document.write( "(1-h)² + (4-k)² = (2-h)² + (5-k)²\r\n" );
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document.write( "1-2h+h² + 16-8k+k² = 4-4h+h² + 25-10k+k²\r\n" );
document.write( "17-2h-8k+h²+k² = 29-4h-10k+h²+k²\r\n" );
document.write( "      17-2h-8k = 29-4h-10k \r\n" );
document.write( "         2h+2k = 12\r\n" );
document.write( "           h+k = 6\r\n" );
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document.write( "Since we are told that the line y = -4x+3 is the perpendicular bisector\r\n" );
document.write( "of the chord EF.  We don't need to know which chord EF is, for ANY line\r\n" );
document.write( "that is the perpendicular bisector of ANY chord must pass through the\r\n" );
document.write( "center of the circle. \r\n" );
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document.write( "So the center (h,k) must lie on the line y = -4x+3, and we \r\n" );
document.write( "can substitute them into the equation of that line and \r\n" );
document.write( "have a true equation:\r\n" );
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document.write( "y = -4x+3\r\n" );
document.write( "k = -4k+3\r\n" );
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document.write( "So we have this system of equations:\r\n" );
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document.write( "h+k=6\r\n" );
document.write( "k = -4h+3\r\n" );
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document.write( "We use substitution:\r\n" );
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document.write( "h+(-4h+3) = 6\r\n" );
document.write( "   h-4h+3 = 6\r\n" );
document.write( "    -3h+3 = 6\r\n" );
document.write( "      -3h = 3\r\n" );
document.write( "        h = -1\r\n" );
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document.write( "k = -4(-1)+3\r\n" );
document.write( "k = 4+3\r\n" );
document.write( "k = 7\r\n" );
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document.write( "So the center is (h,k) = (-1,7)\r\n" );
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document.write( "Use that as a model to solve yours.\r\n" );
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document.write( "Edwin
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