document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=668756>Question 668756</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In circle O, a diameter has endpoints (-5,4) and (3,-6). What's the length of the diamter? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #415820 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=415820\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The length of the diameter is equal to the distance between the two points.\r<BR>\n" );
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document.write( "So use the distance formula to get\r<BR>\n" );
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document.write( "d = sqrt((x2-x1)^2+(y2-y1)^2)\r<BR>\n" );
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document.write( "d = sqrt((3-(-5))^2+(-6-4)^2)\r<BR>\n" );
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document.write( "d = sqrt((3+5)^2+(-6-4)^2)\r<BR>\n" );
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document.write( "d = sqrt((8)^2+(-10)^2)\r<BR>\n" );
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document.write( "d = sqrt(64+100)\r<BR>\n" );
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document.write( "d = sqrt(164)\r<BR>\n" );
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document.write( "d = sqrt(4*41)\r<BR>\n" );
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document.write( "d = sqrt(4)*sqrt(41)\r<BR>\n" );
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document.write( "d = 2*sqrt(41)\r<BR>\n" );
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document.write( "So exact distance between the two points is  2*sqrt(41)  units.\r<BR>\n" );
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document.write( "Using a calculator, the approximate distance is roughly 12.80625  units.\r<BR>\n" );
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document.write( "Therefore, the diameter is exactly <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2%2Asqrt%2841%29 ALIGN=MIDDLE  ALT=\"2%2Asqrt%2841%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > units long. The diameter is approximately  12.80625  units long. </SPAN>\n" );
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