document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=995859>Question 995859</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the equation of a tangent from the point(0,5) to the circle<BR>\n" );
document.write( "x^2 + y^2=16 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #614515 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=614515\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y%5E2=-x%5E2%2B16 ALIGN=MIDDLE  ALT=\"y%5E2=-x%5E2%2B16\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y%5E2=16-x%5E2 ALIGN=MIDDLE  ALT=\"y%5E2=16-x%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=0%2B-+sqrt%2816-x%5E2%29 ALIGN=MIDDLE  ALT=\"y=0%2B-+sqrt%2816-x%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "The tangent from above will meet the upper half;<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=sqrt%2816-x%5E2%29 ALIGN=MIDDLE  ALT=\"y=sqrt%2816-x%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Refer to three different points.  Center of circle is the origin, (0,0);  the given point is (0,5) and is above the circle on the y-axis; and the general point ON the upper branch for the function of the circle,  (x,sqrt(16-x^2)).  There will be two such general points on the circle, one being for x less than 0 and another for x being greater than 0.\r<BR>\n" );
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document.write( "Make a drawing of the description...\r<BR>\n" );
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document.write( "What you want for a line containing (0,5) to be tangent with the circle, is the slopes of line containing  (0,0) and (x,sqrt(16-x^2)); and line containing  (0,5) and (x,sqrt(16-x^2)), to be negative reciprocals of each other.  Think, \"radius\", which will touch the circle at the tangent point(s).\r<BR>\n" );
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document.write( "Formulas for slope, both lines, their product of slopes be negative 1; and then just solve for x.\r<BR>\n" );
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document.write( "After starting the algebra,  the equation to setup is <BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28sqrt%2816-x%5E2%29%2Fx%29%28%28sqrt%2816-x%5E2%29-5%29%2Fx%29=-1 ALIGN=MIDDLE  ALT=\"%28sqrt%2816-x%5E2%29%2Fx%29%28%28sqrt%2816-x%5E2%29-5%29%2Fx%29=-1\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "and omitting the algebraic steps here,<BR>\n" );
document.write( "find that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight%28x=-12%2F5%29 ALIGN=MIDDLE  ALT=\"highlight%28x=-12%2F5%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >  or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight%28x=12%2F5%29 ALIGN=MIDDLE  ALT=\"highlight%28x=12%2F5%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >. </SPAN>\n" );
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