document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1140775>Question 1140775</A>: <I><SPAN STYLE=\"word-break: break-all;\"> (1) A cirlce is defined by the equation x^2-6x+y^2-2y=6.\r<BR>\n" );
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document.write( "(a) Rewrite the equation of the circle in the form:(x-a)^2+(y-b)^2=r^2\r<BR>\n" );
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document.write( "(b) Write down the coordinates of M,the centre of the circle.\r<BR>\n" );
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document.write( "(c) A tangent PQ is drawn from the point P(6;-2) to the circle, with Q the point of contact.Calculate the length of the tangent PQ.\r<BR>\n" );
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document.write( "Please can someone please help me with the question I begg </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #761323 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=761323\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "The answer to part (c) from the other tutor is not for the question that is asked....<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2-6x%2By%5E2-2y=6 ALIGN=MIDDLE  ALT=\"x%5E2-6x%2By%5E2-2y=6\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "Complete the square in x and y to put the equation in standard form<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x-h%29%5E2%2B%28y-k%29%5E2+=+r%5E2 ALIGN=MIDDLE  ALT=\"%28x-h%29%5E2%2B%28y-k%29%5E2+=+r%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%5E2-6x%29%2B%28y%5E2-2y%29+=+6 ALIGN=MIDDLE  ALT=\"%28x%5E2-6x%29%2B%28y%5E2-2y%29+=+6\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%5E2-6x%2B9%29%2B%28y%5E2-2y%2B1%29+=+6%2B9%2B1 ALIGN=MIDDLE  ALT=\"%28x%5E2-6x%2B9%29%2B%28y%5E2-2y%2B1%29+=+6%2B9%2B1\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x-3%29%5E2+%2B+%28y-1%29%5E2+=+4%5E2 ALIGN=MIDDLE  ALT=\"%28x-3%29%5E2+%2B+%28y-1%29%5E2+=+4%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "The center is (3,1); the radius is 4.<br><BR>\n" );
document.write( "We are done with parts (a) and (b) (done correctly by the other tutor).<br><BR>\n" );
document.write( "(c) A tangent PQ is drawn from the point  P(6,-2) to the circle, with Q the point of contact.<br><BR>\n" );
document.write( "Consider the line determined by M and Q.  Let the diameter of the circle contained in that line be AB, with B between M and Q.  Then the rule relating the lengths of a tangent and a secant to a circle tell us<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=PQ%5E2+=+%28PM-MB%29%28PM%2BMA%29 ALIGN=MIDDLE  ALT=\"PQ%5E2+=+%28PM-MB%29%28PM%2BMA%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "MA and MB are radii of the circle; length 4.  MP by the Pythagorean Theorem is 3*sqrt(3).  Then<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=PQ%5E2+=+%283sqrt%283%29-4%29%283sqrt%283%29%2B4%29+=+27-16+=+11 ALIGN=MIDDLE  ALT=\"PQ%5E2+=+%283sqrt%283%29-4%29%283sqrt%283%29%2B4%29+=+27-16+=+11\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "ANSWER: PQ = sqrt(11) </SPAN>\n" );
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