document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=374223>Question 374223</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The two points (-2,4) and (4,2) are the endpoints of the diameter of a circle. Write the equation of the circle in standard form. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #266333 by </B><A HREF=/tutors/aboutme.mpl?userid=robertb><B>robertb(5830)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=robertb><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=robertb><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=266333\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The center of the circle should be midpoints of the two given points: ((-2 + 4)/2, (4+2)/2) = (1, 3).  The radius is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%28%28-2-1%29%5E2+%2B+%284-3%29%5E2%29+=+sqrt%2810%29 ALIGN=MIDDLE  ALT=\"sqrt%28%28-2-1%29%5E2+%2B+%284-3%29%5E2%29+=+sqrt%2810%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  The center-radius form of the circle is then <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x-1%29%5E2+%2B+%28y-3%29%5E2+=+10 ALIGN=MIDDLE  ALT=\"%28x-1%29%5E2+%2B+%28y-3%29%5E2+=+10\"   DISABLED_style= \"max-width:90%; height:auto;\" >. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );