document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=246624>Question 246624</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the equation of the circle with radius 6 that is tangent to both branches of the graph of y=|x|. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #180037 by </B><A HREF=/tutors/aboutme.mpl?userid=dabanfield><B>dabanfield(803)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=dabanfield><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=180037\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's say the center of the circle is C. Draw pependicular lines from C to the two tangent lines created by y = |x|. Let's say the two lines intersect the tangent line at points A and B respectively. Let's say the Origin is point O with coordinates (0,0).\r<BR>\n" );
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document.write( "The figure CAOB is a square with sides of length 6. Triangle CAO is a right isosceles triangle with sides of length 6. So the hypotenuse CO of CAO is sqrt(36+36) = sqrt(72) = 6*sqrt(2). \r<BR>\n" );
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document.write( "The cooredinates of C then are (0,6*sqrt(2)). \r<BR>\n" );
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document.write( "So we have a circle with radius 6 and center C at (0,6sqrt(2).\r<BR>\n" );
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document.write( "Plug these values into the equation of the standard form of a circle:\r<BR>\n" );
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document.write( "(x-h)^2 + (y-k)^2 = r^2 where (h,k) is the center and r is the radius of the circle.     </SPAN>\n" );
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