document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1040626>Question 1040626</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I'm needing some help on this. I submitted earlier, but I didn't get a confirmation it went through:\r<BR>\n" );
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document.write( "I have two points end points on a circle and I need to find the diameter: (–1, 0) and (–9, –8) and I need to find the diameter.\r<BR>\n" );
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document.write( "I've come up with this, but I\"m not sure if it's accurate:\r<BR>\n" );
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document.write( "&#8730;((-1-(-9) )^2+(0-(-8))^2)\r<BR>\n" );
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document.write( "8^2 + 8^2\r<BR>\n" );
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document.write( "64 + 64\r<BR>\n" );
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document.write( "128<BR>\n" );
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document.write( "any help is greatly appreciated! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #655472 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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/cgi-bin/show-face.mpl?user=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=655472\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> It is the square root of 128.  The 8^2+8^2 is correct, but the distance is 8 sqrt (2), which is a simplified term for sqrt (128). This makes intuitive sense, since 8 sqrt (2) is about 11.2, which would be reasonable given the points. </SPAN>\n" );
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