document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=6647>Question 6647</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Dear Sir/Madam, \r<BR>\n" );
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document.write( "I am confronted with the following problem: \r<BR>\n" );
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document.write( "\"Find the equation of the line that is tangent to the circle x^2 + y^2 = 25 at the point P(-3,4).\" \r<BR>\n" );
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document.write( "I did the following:<BR>\n" );
document.write( "1) <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+y%5E2+=+25+-+x%5E2+ ALIGN=MIDDLE  ALT=\"+y%5E2+=+25+-+x%5E2+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "2) <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+y+=+sqrt%2825+-+x%5E2%29+ ALIGN=MIDDLE  ALT=\"+y+=+sqrt%2825+-+x%5E2%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "3) f'(x) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+-2x%2F%282sqrt%2825+-+x%5E2%29%29+ ALIGN=MIDDLE  ALT=\"+-2x%2F%282sqrt%2825+-+x%5E2%29%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "4) f'(-3) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+-6%2F%282sqrt%2816%29%29+=+-6%2F8+=+-3%2F4+ ALIGN=MIDDLE  ALT=\"+-6%2F%282sqrt%2816%29%29+=+-6%2F8+=+-3%2F4+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "5) <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+-3%2F4+=+%28y+-+4%29%2F%28x+%2B+3%29+ ALIGN=MIDDLE  ALT=\"+-3%2F4+=+%28y+-+4%29%2F%28x+%2B+3%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "6) -3(x + 3) = 4(y - 4)<BR>\n" );
document.write( "7) -3x - 9 = 4y - 16<BR>\n" );
document.write( "8) 4y = -3x + 7<BR>\n" );
document.write( "9) <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+y+=+-3x%2F4+%2B+7%2F4+ ALIGN=MIDDLE  ALT=\"+y+=+-3x%2F4+%2B+7%2F4+\"   DISABLED_style= \"max-width:90%; height:auto;\" > which should be the equation of the tangent.\r<BR>\n" );
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document.write( "Instead however, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+y+=+3x%2F4+%2B+25%2F4+ ALIGN=MIDDLE  ALT=\"+y+=+3x%2F4+%2B+25%2F4+\"   DISABLED_style= \"max-width:90%; height:auto;\" > is the equation of the tangent. Why?<BR>\n" );
document.write( "Thanks in advance.<BR>\n" );
document.write( "Regards,<BR>\n" );
document.write( "-Mike </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #3630 by </B><A HREF=/tutors/aboutme.mpl?userid=rapaljer><B>rapaljer(4671)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rapaljer><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=rapaljer><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=3630\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Your problem is in step 2, where you solved for y by taking the square root of both sides of the equation.  You must include a \"+ or -\" symbol in this step, where the plus or the minus is determined by the point that is selected.  What you have is the equation of a circle, and the point of tangency is at (-3,4) placing the point in the second quadrant, since x is negative and y is positive.  Notice that in the second quadrant, the slope of a tangent line to a point on the curve will have a positive slope (by inspection!), so you have to use the plus sign for the slope of the tangent line.  In quadrant IV, where x is positive and y is negative, you also have a positive slope.  In quadrants I and III, the tangent line to a circle will  have a negative slope (again by inspection!).\r<BR>\n" );
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document.write( "I think the rest of what you have done is correct.\r<BR>\n" );
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document.write( "R^2 at SCC </SPAN>\n" );
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