document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=826622>Question 826622</A>: <I><SPAN STYLE=\"word-break: break-all;\"> P(4,2) point on circle x^2+y^2=5x<BR>\n" );
document.write( "If PQ is a diameter of the circle find the coordinates of Q </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #498184 by </B><A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><B>josgarithmetic(39617)</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=498184\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Use the point and see what happens:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%284%29%5E2%2B%282%29%5E2=5%2A4 ALIGN=MIDDLE  ALT=\"%284%29%5E2%2B%282%29%5E2=5%2A4\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=16%2B4=20 ALIGN=MIDDLE  ALT=\"16%2B4=20\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=20=20 ALIGN=MIDDLE  ALT=\"20=20\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Good so far.\r<BR>\n" );
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document.write( "We MUST have radius*radius=20.<BR>\n" );
document.write( "The circle is centered at the origin.  The form of the equations tells that.<BR>\n" );
document.write( "Radius is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2%2Asqrt%285%29 ALIGN=MIDDLE  ALT=\"2%2Asqrt%285%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "The equation for the circle is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2%2By%5E2=20 ALIGN=MIDDLE  ALT=\"x%5E2%2By%5E2=20\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "What have we now?<BR>\n" );
document.write( "The center of the circle is (0,0), based on the equation for it; and the radius is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%2820%29 ALIGN=MIDDLE  ALT=\"sqrt%2820%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  Point P(4,2) is a point on the circle and we WANT the endpoint of the diameter other than P.  \r<BR>\n" );
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document.write( "We also expect the diameter to contain both P and the origin (0,0).  What is the line for these two points?  The both endpoints of the diameter must be on this line!\r<BR>\n" );
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document.write( "Line defined by (0,0) and (4,2):<BR>\n" );
document.write( "slope is (1/2).  Obviously runs through origin.<BR>\n" );
document.write( "Equation for line is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=%281%2F2%29x ALIGN=MIDDLE  ALT=\"y=%281%2F2%29x\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "What points are on this line AND are sqrt(20) distance from the origin (0,0)?  We look for points (x,y), as (x,(1/2)x).  <BR>\n" );
document.write( "USE DISTANCE FORMULA.<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%2820%29=sqrt%28%28x-0%29%5E2%2B%28%281%2F2%29x-0%29%5E2%29 ALIGN=MIDDLE  ALT=\"sqrt%2820%29=sqrt%28%28x-0%29%5E2%2B%28%281%2F2%29x-0%29%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=20=x%5E2%2B%281%2F4%29x%5E2 ALIGN=MIDDLE  ALT=\"20=x%5E2%2B%281%2F4%29x%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%5E2%29%281%2B1%2F4%29=20 ALIGN=MIDDLE  ALT=\"%28x%5E2%29%281%2B1%2F4%29=20\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%285%2F4%29x%5E2=20 ALIGN=MIDDLE  ALT=\"%285%2F4%29x%5E2=20\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2=%284%2F5%2920=16 ALIGN=MIDDLE  ALT=\"x%5E2=%284%2F5%2920=16\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=0%2B-+4 ALIGN=MIDDLE  ALT=\"x=0%2B-+4\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "We already know about x=4 based on our given point on the circle.  We use the equation for the line <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=%281%2F2%29x ALIGN=MIDDLE  ALT=\"y=%281%2F2%29x\"   DISABLED_style= \"max-width:90%; height:auto;\" > to find the OTHER endpoint for the diameter.  Use <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=-4 ALIGN=MIDDLE  ALT=\"x=-4\"   DISABLED_style= \"max-width:90%; height:auto;\" > to find the y value for this endpoint:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=%281%2F2%29%28-4%29 ALIGN=MIDDLE  ALT=\"y=%281%2F2%29%28-4%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=-2 ALIGN=MIDDLE  ALT=\"y=-2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "FINISHED ANSWER: The other endpoint for the diameter at opposite end from (4,2) is (-4,-2).<BR>\n" );
document.write( "DONE!\r<BR>\n" );
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document.write( "-Most of solution above was modified and corrected- </SPAN>\n" );
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