document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=948801>Question 948801</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the equation of the circle that passes through the point (-3,-4) and touches line x-y+7=0 at (-5,2).<BR>\n" );
document.write( "I got x^2+y^2+15x+2y+7+k(3x+y+13) and took the mid points of this equation and substituted to the derieved equation of perpendicular of the given tangent to get the value of k = -11/4. But this doesn't work. What's wrong? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #579113 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=579113\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Examine some of the data carefully.  The given line is equivalent to y=x+7.  You will find that the given point (-5,2) is on this line.  Knowing that should be helpful for you.\r<BR>\n" );
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document.write( "(-5,2)  is a point of tangency.  The line perpendicular to y=x+7 which contains (-5,2) will also contain the center point of the circle.  m=-1 for this line.  y=-x-3 is the line which contains the center point of the circle.  You may well be able to continue from here. </SPAN>\n" );
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