document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1126134>Question 1126134</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the equation of a line tangent to the circle x^2+(y&#8722;3)^2=34  at the point (5, 0).<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #742464 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=742464\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Find the equation of a line tangent to the circle x^2+(y-3)^2=34  at the point (5,0).<BR>\n" );
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document.write( "x^2 + (y-3)^2=34<BR>\n" );
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document.write( "2xdx + 2(y-3)dy = 0<BR>\n" );
document.write( "dy/dx = -x/(y-3) --> the slope at any point on the circle.<BR>\n" );
document.write( "@ (5,0), slope m = 5/3<BR>\n" );
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document.write( "y = (5/3)*(x-5) is the tangent line.<BR>\n" );
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document.write( "Without finding the derivative:<BR>\n" );
document.write( "You can find the slope of the line from the center to the (5,0).<BR>\n" );
document.write( "The slope of the tangent line is the negative inverse.<BR>\n" );
document.write( "etc. </SPAN>\n" );
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