document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=934466>Question 934466</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The point P moves along an arc of a circle with centre E(2,3). The arc of the circle passes through A(-2,0) and B(5,k).<BR>\n" );
document.write( "(a) Find<BR>\n" );
document.write( "(i) the equation of the locus of the point P,<BR>\n" );
document.write( "(ii) the values of k.<BR>\n" );
document.write( "(b) The tangent of the circle at the point A intersects the y-axis at the point Q. Find the area of triangle OAQ where O is 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 #567808 by </B><A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><B>josgarithmetic(39620)</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=567808\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The length EA can be found with Distance Formula.  This will be the length of the radius.  k can then be found because EA=EB.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=EA=sqrt%28%282-%28-2%29%29%5E2%2B%283-0%29%5E2%29 ALIGN=MIDDLE  ALT=\"EA=sqrt%28%282-%28-2%29%29%5E2%2B%283-0%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=sqrt%28%284%29%5E2%2B%283%29%5E2%29 ALIGN=MIDDLE  ALT=\"sqrt%28%284%29%5E2%2B%283%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=sqrt%28%2816%29%5E2%2B%289%29%5E2%29 ALIGN=MIDDLE  ALT=\"sqrt%28%2816%29%5E2%2B%289%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=sqrt%2825%29 ALIGN=MIDDLE  ALT=\"sqrt%2825%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight_green%285=r%29 ALIGN=MIDDLE  ALT=\"highlight_green%285=r%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Knowing E(2,3) is the center, according to standard form equation of a circle,<BR>\n" );
document.write( "The equation for this circle is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight%28%28x-2%29%5E2%2B%28y-3%29%5E2=25%29 ALIGN=MIDDLE  ALT=\"highlight%28%28x-2%29%5E2%2B%28y-3%29%5E2=25%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "Part (b) is more work to solve.<BR>\n" );
document.write( "Find slope of EA.<BR>\n" );
document.write( "Form the negative reciprocal of that slope.<BR>\n" );
document.write( "The line tangent at point A and perpendicular to the segment EA can be found  (its equation).<BR>\n" );
document.write( "Using this new equation, determine its y-intercept.<BR>\n" );
document.write( "I have not worked through this part completely.  Your triangle might or might not be a Right triangle.\r<BR>\n" );
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