document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=258001>Question 258001</A>: <I><SPAN STYLE=\"word-break: break-all;\"> y-9=0<BR>\n" );
document.write( "x^2+y^2-24x-10y+153=0 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #189873 by </B><A HREF=/tutors/aboutme.mpl?userid=Greenfinch><B>Greenfinch(383)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Greenfinch><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=Greenfinch><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=189873\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x^2 - 24x + y^2 - 10y + 153 = 0 is a circle with centre at 12,5<BR>\n" );
document.write( "So rearranging<BR>\n" );
document.write( "(x^2 - 24x + 144) + (y^2 -  10y + 25) + 153 - 169 = 0 completing the square<BR>\n" );
document.write( "(x - 12)^2 + (y - 5)^2 = 16 = 4^2 Centre is there at 12, 5 and the radius is 4<BR>\n" );
document.write( "Combining with the line y = 9 gives<BR>\n" );
document.write( "(x - 12)^2 + (y - 5)^2 - 16 =0 and y = 9<BR>\n" );
document.write( "(x - 12)^2 + 4^2 - 16 = 0<BR>\n" );
document.write( "(x - 12) ^2 = 0 <BR>\n" );
document.write( "x = 12 twice, so y = 9 is a tangent touching the circle at x = 12<BR>\n" );
document.write( " </SPAN>\n" );
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