document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=143914>Question 143914</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The center of circle P  is at  (8,6) and (13, 6 - sqrt39) is a point on  the  circle. If circle  P is inscribed in square ABCD, what is the area of ABCD? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #105233 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=105233\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Given the center of a circle and a point on the circle, you can use the distance formula <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=d=sqrt%28+%28x%5B1%5D-x%5B2%5D%29%5E2+%2B+%28y%5B1%5D-y%5B2%5D%29%5E2%29 ALIGN=MIDDLE  ALT=\"d=sqrt%28+%28x%5B1%5D-x%5B2%5D%29%5E2+%2B+%28y%5B1%5D-y%5B2%5D%29%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > to calculate the radius.  The diameter of a circle inscribed in a square is equal to the side length of the square.  So, take the radius that you calculated, double it, and then square the result for the area of the square. </SPAN>\n" );
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