document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1085760>Question 1085760</A>: <I><SPAN STYLE=\"word-break: break-all;\"> what is the equation of this line?\r<BR>\n" );
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document.write( "question: find the equations of the line described.<BR>\n" );
document.write( "tangent lines to x²+y²-18x-2y+48=0 at points where x=4\r<BR>\n" );
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document.write( "im solving it using x²+y²-18y-2y+48=0<BR>\n" );
document.write( "at x=4\r<BR>\n" );
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document.write( "my solution (trying)\r<BR>\n" );
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document.write( "P(4,y)\r<BR>\n" );
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document.write( "(4)²+y²-18(4)-2y+48=0<BR>\n" );
document.write( "16+y²-72-2y+48=0<BR>\n" );
document.write( "=y²+2y<BR>\n" );
document.write( "=y(y+2)=0<BR>\n" );
document.write( "y=0 | y+2=0 y=-2\r<BR>\n" );
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document.write( "is it right? \r<BR>\n" );
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document.write( "P(4,0) Q(4,-2) is it right?\r<BR>\n" );
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document.write( "x²+y²-18x-2y+48=0<BR>\n" );
document.write( "(x²-18x)+(y²-2y)= -48 \r<BR>\n" );
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document.write( "is it right? now i dont know whats next. <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 #699856 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><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=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=699856\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Your approach is correct to find the y-coordinates of the intersection points, but there is an error in your arithmetic.<BR>\n" );
document.write( "When you substitute x=4 into the equation of the circle, you should get<BR>\n" );
document.write( "y^2 - 2y - 8 = 0 and this factors as (y-4)(y+2) = 0<BR>\n" );
document.write( "Thus the two intersection points are (4,4) and (4,-2)<BR>\n" );
document.write( "We use implicit differentiation to find dy/dx, which is the slope of the tangent line to the circle:<BR>\n" );
document.write( "2xdx + 2ydy - 18dx -2dy = 0<BR>\n" );
document.write( "-2xdx = (2y-18)dy -> dy/dx = (9-x)/(y-1)<BR>\n" );
document.write( "We want to find dy/dx at the points (4,4) and (4,-2)<BR>\n" );
document.write( "Substituting the values we get dy/dx @ (4,4) = 5/3 and dy/dx @ (4,-2) = -5/3<BR>\n" );
document.write( "To find the equations for the lines, we use the point-slope form:<BR>\n" );
document.write( "Line 1: y-4 = (5/3)(x-4) -> y = (5/3)x - 8/3<BR>\n" );
document.write( "Line 2: y+2 = (-5/3)(x-4) -> y = (-5/3)x + 14/3 </SPAN>\n" );
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