document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1079796>Question 1079796</A>: <I><SPAN STYLE=\"word-break: break-all;\"> he general form of the equation of a circle is x2+y2+2x&#8722;6y+1=0.\r<BR>\n" );
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document.write( "What are the coordinates of the center of the circle? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #694022 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=694022\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> rewrite in form of (x-h)^2+(y-k)^2=r^2<BR>\n" );
document.write( "x^2+2x+y^2-6y=-1<BR>\n" );
document.write( "complete the square of both<BR>\n" );
document.write( "x^2+2x+1+y^2-6y+9=9, after adding 20 to both sides<BR>\n" );
document.write( "(x+1)^2+(y-3)^2=3^2<BR>\n" );
document.write( "The center is the additive inverse of both constants, so it is at (-1,3) and the radius is 3, the square root of 9.<BR>\n" );
document.write( "The center is at (-1, 3)<BR>\n" );
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