document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=206555>Question 206555</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do you find the standard form of the equation of a circle, with a center (3,-1) through the point (-5,1)? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #514813 by </B><A HREF=/tutors/aboutme.mpl?userid=tientrinh><B>tientrinh(1)</B></A><img src=\"/images/stars/iconNewId_16x16.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=tientrinh><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=514813\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Find the distance to the point, that's the radius. By the formula d= &#8730;(x2-x1)^2+ (y2-y1)^2  <BR>\n" );
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document.write( "d= &#8730;(3+5)^2+ (-1-1)^2  <BR>\n" );
document.write( "d= &#8730;64+4 = &#8730;68 (or =r)<BR>\n" );
document.write( "r^2=68\r<BR>\n" );
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document.write( "Then it's (x-h)^2 + (y-k)^2 = r^2 (h,k) is the center<BR>\n" );
document.write( "(x-3)^2 + (y+1)^2 = 68 </SPAN>\n" );
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