document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=366780>Question 366780</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What is the sum of the coordinates of the center of the circle whose equation is\r<BR>\n" );
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document.write( "x^2 + y^2 - 6x - 10y + 30 = 0 ? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #261457 by </B><A HREF=/tutors/aboutme.mpl?userid=mananth><B>mananth(16946)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mananth><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=mananth><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=261457\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x^2 + y^2 - 6x - 10y + 30 = 0<BR>\n" );
document.write( "..<BR>\n" );
document.write( "x^2-6x +y^2-10y+30=0<BR>\n" );
document.write( "(x^2-6x+9)-9+(y^2-10y+25)-25+30=0<BR>\n" );
document.write( "(x-3)^2+(y-5)^2-9-25+30=0<BR>\n" );
document.write( "(x-3)^2+(y-5)^2=9+25-30<BR>\n" );
document.write( "(x-3)^2+(y-5)^2=4<BR>\n" );
document.write( "Center = (3,5) <BR>\n" );
document.write( "sum = 8<BR>\n" );
document.write( "...<BR>\n" );
document.write( "m.ananth@hotmail.ca<BR>\n" );
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