document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=2702>Question 2702</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please do you have a simple formula for finding the centre of a 2 dimensinal circle if I know 3 points on this circle? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #1260 by </B><A HREF=/tutors/aboutme.mpl?userid=xcentaur><B>xcentaur(357)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=xcentaur><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=1260\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the equation of the circle be\r<BR>\n" );
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document.write( "     (x-h)^2 + (y-k)^2 = r^2,\r<BR>\n" );
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document.write( "and substitute the three known points, getting 3 equations in 3 <BR>\n" );
document.write( "unknowns h, k, and r:\r<BR>\n" );
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document.write( "     (x1-h)^2 + (y1-k)^2 = r^2<BR>\n" );
document.write( "     (x2-h)^2 + (y2-k)^2 = r^2<BR>\n" );
document.write( "     (x3-h)^2 + (y3-k)^2 = r^2\r<BR>\n" );
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document.write( "which you can solve simultaneously. First subtract the third equation <BR>\n" );
document.write( "from the other two, thus eliminating r^2, h^2, and k^2. That will <BR>\n" );
document.write( "leave you with 2 simultaneous linear equations in h and k to solve. <BR>\n" );
document.write( "This you can do as long as the 3 points are not collinear. Then those <BR>\n" );
document.write( "values of h and k can be used in the first equation to find the <BR>\n" );
document.write( "radius:\r<BR>\n" );
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document.write( "     r = sqrt[(x1-h)^2 + (y1-k)^2].\r<BR>\n" );
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document.write( "Example: Suppose a circle passes through the points (4,1), (-3,7), and <BR>\n" );
document.write( "(5,-2). Then we know that:\r<BR>\n" );
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document.write( "     (h-4)^2 + (k-1)^2 = r^2<BR>\n" );
document.write( "     (h+3)^2 + (k-7)^2 = r^2<BR>\n" );
document.write( "     (h-5)^2 + (k+2)^2 = r^2\r<BR>\n" );
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document.write( "Subtracting the first from the other two, you get:\r<BR>\n" );
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document.write( "     (h+3)^2 - (h-4)^2 + (k-7)^2 - (k-1)^2 = 0,<BR>\n" );
document.write( "     (h-5)^2 - (h-4)^2 + (k+2)^2 - (k-1)^2 = 0,\r<BR>\n" );
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document.write( "     h^2 + 6h + 9 - h^2 + 8h - 16 + k^2 - 14k + 49 - k^2 + 2k - 1 = 0<BR>\n" );
document.write( "     h^2 - 10h + 25 - h^2 + 8h - 16 + k^2 + 4k + 4 - k^2 + 2k - 1 = 0\r<BR>\n" );
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document.write( "     14h - 12k + 41 = 0<BR>\n" );
document.write( "     -2h +  6k + 12 = 0\r<BR>\n" );
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document.write( "     10h +  65 = 0<BR>\n" );
document.write( "     30h + 125 = 0\r<BR>\n" );
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document.write( "     h = -13/2<BR>\n" );
document.write( "     k = -25/6\r<BR>\n" );
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document.write( "Then\r<BR>\n" );
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document.write( "     r = sqrt[(4+13/2)^2 + (1+25/6)^2]<BR>\n" );
document.write( "       = sqrt[4930]/6\r<BR>\n" );
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document.write( "Therefore the equation of the circle is:\r<BR>\n" );
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document.write( "     (x+13/2)^2 + (y+25/6)^2 = 4930/36<BR>\n" );
document.write( "<P><BR>\n" );
document.write( "I learnt this very same problem from a site called http://www.mathforum.org/<BR>\n" );
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document.write( "Hope this helps,<BR>\n" );
document.write( "Best of luck </SPAN>\n" );
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