document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=993811>Question 993811</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find an equation for the line that is tangent to the circle x^2+y^2=169 at the point (5,12).<BR>\n" );
document.write( "I need help beind reminded of the equation of a circle and equations related to this, and also a step-by-step explantion to help me to understand how to solve this typ of problems and others efficiently. Thank you so much! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #612995 by </B><A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><B>josgarithmetic(39618)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=612995\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Center of circle on the Origin,  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2%2By%5E2=r%5E2 ALIGN=MIDDLE  ALT=\"x%5E2%2By%5E2=r%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >, and for this, radius is r.<BR>\n" );
document.write( "You have <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r%5E2=169=13%5E2 ALIGN=MIDDLE  ALT=\"r%5E2=169=13%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so the radius <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r=13 ALIGN=MIDDLE  ALT=\"r=13\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "The point  ON THE CIRCLE, (5,12), is part of a tangent line which passes through this point.  This means, you want to find an equation for this line and this line TOUCHES the circle at this point; and it is perpendicular to the line which contains  this point (5,12) and the Origin (which is center of your circle.)<BR>\n" );
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document.write( "Work to understand that discussion before continuing.\r<BR>\n" );
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document.write( "What is the line containing the circle's center (0,0) and the given point (5,12)?  You should find just intuitively this is  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=%2812%2F5%29x ALIGN=MIDDLE  ALT=\"y=%2812%2F5%29x\"   DISABLED_style= \"max-width:90%; height:auto;\" >.   The y-intercept and the x-intercept both 0.\r<BR>\n" );
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document.write( "What is the equation for the line PERPENDICULAR to <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=%2812%2F5%29x ALIGN=MIDDLE  ALT=\"y=%2812%2F5%29x\"   DISABLED_style= \"max-width:90%; height:auto;\" >  and contains the point  (5,12)?  For perpendicularity, its slope must be negative reciprocal of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=12%2F5 ALIGN=MIDDLE  ALT=\"12%2F5\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so this slope needed will be <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-5%2F12 ALIGN=MIDDLE  ALT=\"-5%2F12\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  You can use the point-slope equation form  (for convenience if you are comfortable with it), and plug-in the needed slope and included point (5,12):<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight%28y-12=-%285%2F12%29%28x-5%29%29 ALIGN=MIDDLE  ALT=\"highlight%28y-12=-%285%2F12%29%28x-5%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Use simple algebra if you want this equation in standard form or in slope-intercept form.\r<BR>\n" );
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document.write( "Try to make a sketch or a graph on your own to help analyze the problem description. </SPAN>\n" );
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