document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1048633>Question 1048633</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Determine the value(s) of k such that the circle x^2+(y-6)^2 = 36  and the parabola x^2 = 4ky  will intersect only at the origin. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #664336 by </B><A HREF=/tutors/aboutme.mpl?userid=robertb><B>robertb(5830)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=robertb><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=robertb><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=664336\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> From the system <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2%2B%28y-6%29%5E2+=+36 ALIGN=MIDDLE  ALT=\"x%5E2%2B%28y-6%29%5E2+=+36\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+x%5E2+=+4ky+ ALIGN=MIDDLE  ALT=\"+x%5E2+=+4ky+\"   DISABLED_style= \"max-width:90%; height:auto;\" > we fast-forward to the point where, after substitution and a little algebra, we get\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y%5E2%2B%284k-12%29y+=+0 ALIGN=MIDDLE  ALT=\"y%5E2%2B%284k-12%29y+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "Case 1. \r<BR>\n" );
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document.write( "We know beforehand that the two curves intersect at (0,0). To ensure that the above equation has at least one solution, let the discriminant \r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2-4ac+=+%284k-12%29%5E2+-+4%2A1%2A0+=+%284k-12%29%5E2+%3E=+0 ALIGN=MIDDLE  ALT=\"b%5E2-4ac+=+%284k-12%29%5E2+-+4%2A1%2A0+=+%284k-12%29%5E2+%3E=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "===> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=4k-12+%3E=+0 ALIGN=MIDDLE  ALT=\"4k-12+%3E=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >  ===> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=k+%3E=+3 ALIGN=MIDDLE  ALT=\"k+%3E=+3\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "Now when k = 3, the circle will \"fit\" snugly on the parabola at the origin, hence intersect only at that point.  \r<BR>\n" );
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document.write( "We focus on the instance k > 3.\r<BR>\n" );
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document.write( "===> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y%5E2%2B%284k-12%29y+=+y%28y%2B%284k-12%29%29+=+0 ALIGN=MIDDLE  ALT=\"y%5E2%2B%284k-12%29y+=+y%28y%2B%284k-12%29%29+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >  ===> y = 0, -(4k-12).\r<BR>\n" );
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document.write( "But <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+-%284k-12%29+%3C+0 ALIGN=MIDDLE  ALT=\"y+=+-%284k-12%29+%3C+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  If k > 3 are plugged into <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+x%5E2+=+4ky+ ALIGN=MIDDLE  ALT=\"+x%5E2+=+4ky+\"   DISABLED_style= \"max-width:90%; height:auto;\" >, we will be solving for <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+x%5E2+=+-4k%284k-12%29+%3C+0+ ALIGN=MIDDLE  ALT=\"+x%5E2+=+-4k%284k-12%29+%3C+0+\"   DISABLED_style= \"max-width:90%; height:auto;\" >, a contradiction, hence not producing new solutions other than the origin.\r<BR>\n" );
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document.write( "Therefore, for k in [3, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=infinity ALIGN=MIDDLE  ALT=\"infinity\"   DISABLED_style= \"max-width:90%; height:auto;\" >), the parabola <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+x%5E2+=+4ky+ ALIGN=MIDDLE  ALT=\"+x%5E2+=+4ky+\"   DISABLED_style= \"max-width:90%; height:auto;\" > will intersect the circle only at the origin.\r<BR>\n" );
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document.write( "Case 2.  For k in (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-infinity ALIGN=MIDDLE  ALT=\"-infinity\"   DISABLED_style= \"max-width:90%; height:auto;\" >, 0]:\r<BR>\n" );
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document.write( "If k = 0, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+x%5E2+=+4ky+ ALIGN=MIDDLE  ALT=\"+x%5E2+=+4ky+\"   DISABLED_style= \"max-width:90%; height:auto;\" > becomes <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2+=+0 ALIGN=MIDDLE  ALT=\"x%5E2+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >, or the vertical line x = 0, which is not a parabola, so k = 0 is excluded.  <BR>\n" );
document.write( "For k < 0, the parabola opens downward (can accept only negative y values), hence it will automatically intersect the circle only at the origin.\r<BR>\n" );
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document.write( "Case 3.  0 < k < 3:\r<BR>\n" );
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document.write( "It can be shown that for 3/2 < k < 3, the parabola will intersect the circle at two points of equal level at the LOWER semicircular part.<BR>\n" );
document.write( "For k = 3/2, the parabola intersects at the points (-6,6) and (6,6) (the diametrical points!).<BR>\n" );
document.write( "For 0 < k < 3/2, the parabola will intersect the circle at two points of equal level at the UPPER semicircular part.\r<BR>\n" );
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document.write( "Therefore for Case 3, there are exactly 3 points of intersection, including the origin.\r<BR>\n" );
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document.write( "The final answer, therefore, is that k should be in (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-infinity ALIGN=MIDDLE  ALT=\"-infinity\"   DISABLED_style= \"max-width:90%; height:auto;\" >,0)&#8746;[3, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=infinity ALIGN=MIDDLE  ALT=\"infinity\"   DISABLED_style= \"max-width:90%; height:auto;\" >), for the origin to be the ONLY intersection point.	\r<BR>\n" );
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