document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=877020>Question 877020</A>: <I><SPAN STYLE=\"word-break: break-all;\"> WHAT IS THE EQUATION OF A CIRCLE TOUCHING THE LINES <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+x+-+3y+-+11+=+0+ ALIGN=MIDDLE  ALT=\"+x+-+3y+-+11+=+0+\"   DISABLED_style= \"max-width:90%; height:auto;\" > AND <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+3x+-+y+-+9+=+0+ ALIGN=MIDDLE  ALT=\"+3x+-+y+-+9+=+0+\"   DISABLED_style= \"max-width:90%; height:auto;\" > AND, HAVING ITS CENTER ON THE LINE <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+x+%2B+2y+%2B+19+=+0+ ALIGN=MIDDLE  ALT=\"+x+%2B+2y+%2B+19+=+0+\"   DISABLED_style= \"max-width:90%; height:auto;\" > ? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #529113 by </B><A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><B>josgarithmetic(39630)</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=529113\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Take a viewpoint from above the plane containing the lines, with the three lines graphed.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28300%2C300%2C-12%2C12%2C-12%2C12%2Cx%2F3-11%2F3%2C3x-9%2C-x%2F2-19%2F2%29 ALIGN=MIDDLE  ALT=\"graph%28300%2C300%2C-12%2C12%2C-12%2C12%2Cx%2F3-11%2F3%2C3x-9%2C-x%2F2-19%2F2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "The blue line is the one which contains the center of the circle.\r<BR>\n" );
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document.write( "This will be messy, but you might use Distance formula.  The general point of the center of circle is (x, -x/2-19/2).  General point of one of the lines is (x, x/3-11/3) and of the other line is (x, 3x-9). \r<BR>\n" );
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document.write( "The center point is equadistant from the two other general points.\r<BR>\n" );
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document.write( "This is what becomes messy:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%28%28%28-x%2F2-19%2F2%29-%28x%2F3-11%2F3%29%29%5E2%2B%28x-x%29%5E2%29=sqrt%28%28%28-x%2F2-19%2F2%29-%283x-9%29%29%5E2%2B%28x-x%29%5E2%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "That should be done on paper and solve for x.  This will let you find y, and these are coordinates for the center of the circle.\r<BR>\n" );
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document.write( "The distance formula equation shown above seems to simplify to <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=416x%5E2-224x-1216=0 ALIGN=MIDDLE  ALT=\"416x%5E2-224x-1216=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  Further reduces through factorizations GCF of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2%5E5 ALIGN=MIDDLE  ALT=\"2%5E5\"   DISABLED_style= \"max-width:90%; height:auto;\" >, to <BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=highlight_green%2813x%5E2-7x-38=0%29 ALIGN=MIDDLE  ALT=\"highlight_green%2813x%5E2-7x-38=0%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >. </SPAN>\n" );
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