document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=864894>Question 864894</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The line with the equation y=mx is tangent to the circle with centre (-8,0) and radius 4 at the point P(x,y).\r<BR>\n" );
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document.write( "a)find coordinates of P in terms of m\r<BR>\n" );
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document.write( "b)Show that m = +- sqrt (3)/ 3 and hence find the coordinates of P </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #521353 by </B><A HREF=/tutors/aboutme.mpl?userid=richwmiller><B>richwmiller(17219)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richwmiller><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=richwmiller><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=521353\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> (x+8)^2+y^2=16,\r<BR>\n" );
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document.write( "(x+8)^2+y^2=16, y=mx<BR>\n" );
document.write( "m = 1/sqrt(3)<BR>\n" );
document.write( "m =-1/sqrt(3)\r<BR>\n" );
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document.write( "Multiply by Sqrt(3)/Sqrt(3) to get whole number in denominator<BR>\n" );
document.write( "and get +-sqrt(3)/3\r<BR>\n" );
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document.write( "+-1/sqrt(3)=+-sqrt(3)/3<BR>\n" );
document.write( "y=+-sqrt(3)x/3<BR>\n" );
document.write( "x=-6,   y =2sqrt(3)<BR>\n" );
document.write( "P=(-6,2sqrt(3)) </SPAN>\n" );
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