document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1082354>Question 1082354</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the equation of a line that is tangent to the circle x^2 + y^2 - 8x - 8y + 7 = 0 at the point (1,0).<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 #696401 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=696401\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  Find the  equation of a line that is tangent to the circle x^2 + y^2 - 8x - 8y + 7 = 0 at the point (1,0).<BR>\n" );
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document.write( "Take the derivative to find the slope at every point.<BR>\n" );
document.write( "2x + 2y*y'-8- 8y<BR>\n" );
document.write( "= 0<BR>\n" );
document.write( "y' = (8-2x)/(2y-8)<BR>\n" );
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document.write( "Slope at (1,0)<BR>\n" );
document.write( "slope = (8-2)/(0-8) = 6/(-8) = -3/4<BR>\n" );
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document.write( "Form of line:: y = mx + b<BR>\n" );
document.write( "Solve for \"b\"::<BR>\n" );
document.write( "0 = (-3/4)1 + b<BR>\n" );
document.write( "b = (3/4)<BR>\n" );
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document.write( "Equation:: y = (-3/4)x + (3/4)<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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