document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1084215>Question 1084215</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In the (x,y) coordinate plane, which of the following is an equation of the circle having the points (3,-3) and (-5,5) as endpoints of a diamteter? <BR>\n" );
document.write( "F) (x+1)^2 + (y-1)^2 = sqrt 32 G) (x+1)^2 + (y+1)^2 = sqrt 32 H) (x-1)^2 + (y+1)^2 = sqrt 32<BR>\n" );
document.write( "J) (x-1)^2 + (y+1)^2 = 32  K) (x+1)^2 + (y-1)^2 = 32 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #698286 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=698286\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Need to know the center and the radius.<BR>\n" );
document.write( "Midpoint of diameter is center, which is midpoint formula, which gives (-1, 1) and the radius is the distance from the center to either end of the diameter.  That is the distance formula, sqrt ((x2-x1)^2+(y2-y1)^2)= sqrt (16+16)=sqrt (32)<BR>\n" );
document.write( "so r^2=32.  The equation of a circle is x^2+y^2=r^2.  So once one has the sqrt of the radius (here sqrt (32)), it has to be squared (32) to be part of the equation.  <BR>\n" );
document.write( "(x+1)^2+(y-1)^2=32<BR>\n" );
document.write( "K. </SPAN>\n" );
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